The Joule is an attempt to quantify energy, the idea of "effort", or the ability to exert effort, to do "work". The metre is easy to describe, you just hold your hands so far apart, and everybody now knows what a metre is. Heat and mass are similarly easy, because they can be demonstrated empirically, but energy (of which the Joule is a measure) is difficult.
Energy seems to be at the root of all of existence, everything that exists is some manifestation of it. "Work" is the transformation of energy from one of these manifest forms to another. A large chunk of scientific endeavour over the last few hundred years has been to quantify it, and the result is a handful of algebraic formulae that tell you how many joules of energy comprise its different manifest forms.
For example, the number of Joules of energy "contained" in a photon of light:
$$ E = \frac{hc}{\lambda}$$
The energy, in Joules, possessed by a mass due solely to its motion in space (kinetic energy):
$$ E = \frac{1}{2}mv^2 $$
The number of Joules "contained" by an object at rest, due to its mass alone:
$$ E = mc^2 $$
These formulae do not describe any real, tangible thing, but rather they yield abstract values which reveal a remarkable truth about energy - that it's always conserved. That is in any closed system or experiment, when you use them to find all values of all forms of energy in that system, the sum total never changes, even if the forms do.
This is the true importance and meaning of energy (Joules), not that it represents anything on its own (because what it represents can be anything and everything, in fact), only that we are able to predict that if something loses X Joules, then something else gains X Joules.
In other words, conservation of energy enables us to very precisely predict and control the transfer and transformation of energy between various forms, and the meaning of a "Joule" depends on the form it currently holds.
The best I can do to illustrate a "Joule" is to describe its manifest effect in the context of things you are familiar with, so I'll do that for a few examples.
When you lift a mass 1kg vertically upwards through a distance of 1m, the mass has \$ E = mgh = 1 \text{ kg} \times 9.81 \text{ ms}^\text{-2} \times 1\text{ m} = 9.81 \text{ J} \$ of energy more than it had before, in the form of gravitational potential energy. If the muscles you used to lift it were 100% efficient, they would have \$9.81 \text{ J} \$ less chemical potential energy. In other words, they did 9.81J of work, lifting the mass. They are not 100% efficient, so other work will also have been done, mostly in the form of heating.
It will take 4200J of energy to raise the temperature of 1kg of water by 1°C. If you wish to boil a litre of water from 25°C, you will need to "deliver" \$ 4200 \text{ Jkg}^{-1}{K}^{-1} \times 1\text{ kg} \times (100-25)\text{K} = 315\text{kJ}\$ of energy to the water. We would say you need to do 315 kJ of work. A 1kW kettle would take at least 315s to complete the job, certainly longer due to inefficiencies.
A photon of red light has about \$ E = \frac{hc}{\lambda} = \frac{6.6\times 10^{-34}\times 3.0\times 10^8}{700\times 10^{-9}} = 2.8\times 10^{-19}\text{ J}\$ of energy. A 50% efficient 5mW red laser delivers 0.0025J of light energy each second, which would be \$ \frac{0.0025}{2.8\times 10^{-19}} = 8.9\times 10^{15}\$ photons per second.
You asked "how is a Joule measured?". Again, that depends on the form of the Joule of energy. If the form is light, and you know the wavelength of the light, you can literally count photons, and apply the formulae above. Or, perhaps, you could absorb the light into a black light-absorbent material, and measure the rise in temperature of the material. If you want to know how many Joules of energy are stored in a battery, you could make the battery turn a motor to lift a mass. There are other, probably better ways, but you get the idea.
Now onto the Volt. A better analogy than water is gravity, and a ball rolling downhill. At the top of a hill, a ball has more gravitational potential energy that it would at the bottom, due to the gravitational field of the planet. It accelerates downwards because gravity exerts a force upon it. As it accelerates downhill, it loses gravitational potential energy, and gains kinetic energy (velocity - remember, conservation of energy). On its way, it collides and interacts with whatever obstacles it encounters, transferring some of its newly gained kinetic energy to the obstacle. Due to obstacles and other forces at work, the ball is not free to continue accelerating, rather it achieves a sort of "terminal velocity", and by the time it reaches the bottom of the hill, it has expended all the potential energy it began with, arriving with almost no kinetic energy, at a place where it has no potential energy either.
In other words, if the ball starts with 100J of potential energy at the top, almost all of that energy is delivered to the various obstacles on its path, and when it reaches the bottom with (close to) 0J of kinetic energy, and 0J of potential energy.
The potential energy it has at any point on its journey represents the "voltage" at that point. At the top it started with 100J. Half way down, the ball had 50J of potential energy, so we could say the "voltage" there is 50 (the units are messed up in this analogy, but you get the point). At the bottom, the potential (voltage) is zero. It's important to note that you could called the potential energy at the top 100J, 0J or 1,000,000J, as long as it has 100J less at the bottom (in which case you'd say that the "voltage" at the bottom is 0, -100 or 999900 respectively) - everything is relative, and we usually talk in terms of potential differences, not absolute potentials.
For electrons, if the potential at the point where the charge currently finds itself (point A) has a "voltage" (AKA "potential") which is 10V higher than charges at point B, then any charge at point A will have 10eV of potential energy more than a charge at point B. On a journey from A to B, that charge will start the journey with 10eV of potential energy, and end with none. All of that energy will have been donated to whatever obstacles it encountered on the way, making light, heat, motion etc.
The electron charge is \$ q=1.9\times 10^{-19}C \$, meaning that you require \$ \frac{1}{1.9\times 10^{-19}} = 6.2\times 10^{18}\$ electrons to have 1 Coulomb of charge. The electron-volt is a unit of energy equal to \$ 1eV = qJ \$ Consequently, you would need \$ 6.2\times 10^{18} \$ electrons (a total of 1C) to make that journey to deliver all 10J of energy to whatever they flow through. Alternatively, twice that number making half the journey would also deliver ("dissipate") 10J of energy, so it's not necessary for a charge to make the entire journey, only that lots of charges to travel some fraction of the distance.
Charges always experience a force in the direction that accelerates them to a place of lower potential energy. If there's no potential difference (AKA EMF, voltage) explicitly applied across a circuit or conductor, charges settle into a position of equilibrium, where they all have equal forces left and right, above below etc, with no net force to accelerate them away from that equilibrium. Often that's a uniform distribution of charges throughout the conductor, in which no charge has a higher or lower potential energy than any other. In other words, the potential difference (voltage) between any two points in the conductor is zero.
The only means by which a charge would feel compelled to move, is if it finds itself in a region where charge density to one side is greater than density on the other, resulting in a repulsion away from the higher density region, and an attraction toward the lower density region. Therefore, to compel charges to move, it is necessary to artificially introduce a non-uniformity, which we can do with cells or magnetic fields.
With a cell, we charge it by doing work to ionise molecules, and separate electrons from their parent atoms' nuclei. By grouping electrons to one side, and leaving the now-positively charged nuclei on the other, we have produced a charge imbalance, a dense grouping of negative charges physically separated from a dense group of positive charges, which have no conductive path via which to travel and redress the imbalance. The charges on each side now have a large potential energy. For example, in a 1.5V cell, each charge has 1.5eV of electrical potential energy, experiences a strong force towards its opposite partner, but has no way to travel there.
By connecting the electrodes of the charged cell to an external conductor or circuit, the charge density imbalance in the cell repels/attracts charges in the external conductors, suddenly endowing them also with 1.5eV each. In this way the connected circuit now also has a 1.5V potential difference across it.
The electric field due to the charge separation in the cell is now forcing charges in the connected circuit to move. Starting at the cell, charges with 1.5eV of potential energy can travel out and around towards a place of lower potential energy, eventually returning to the cell with none, the recipient of all that energy being whatever it travelled through to get back to the cell.
I do not agree that charges have a set speed, which does not increase as voltage is raised. Doubling voltage doubles electron drift velocity. If the conductor is purely resistive, then not only will velocity double (causing a doubling in coulombs-per-second current), but the starting potential energy of each charge is doubled. This means that energy-per-second dissipation (power) in the conductor increases by a factor of four, giving rise to the square term in \$ P = \frac{V^2}{R} \$
