First, what is a transformation? A transform is a mapping of a function in one domain into another domain. To go from one domain to another, you need basis-functions (basis-vectors).
In Fourier transform, the basis-functions are complex exponentials; \$e^{-j\omega t} = \cos(\omega t) - j\sin(\omega t)\$
In Laplace transform, the basis-functions are the product of complex exponentials and decaying exponentials; \$e^{-\sigma t}e^{-j\omega t} = e^{-(\sigma+j\omega) t} = e^{-st}\$.
Fourier transform: \$F(j\omega) = \displaystyle\int_{-\infty}^\infty e^{-j\omega t}f(t) \: \text{d} t \$
Laplace transform: \$F(s) = \displaystyle\int_{-\infty}^\infty e^{-st}f(t) \: \text{d} t \$
Think about the product inside the integral (\$e^{-j\omega t}f(t)) \$ as an inner product (dot product) between two vectors. A dot product is large, if the two vectors are similar to each other. A dot product is small, if the two vectors are dissimilar.
Consider the Fourier transform first. Evaluating \$e^{-j\omega t}f(t)\$ we investigate how similar \$f(t)\$ is to \$\big(\cos(\omega t) - j\sin(\omega t)\big) \$ for a single frequency.
Evaluating \$\displaystyle\int_{-\infty}^\infty e^{-j\omega t}f(t) \: \text{d} t \$ you get a function telling you how similar \$f(t)\$ is to \$\big(\cos(\omega t) - j\sin(\omega t)\big) \$ for any frequency. We call this function \$F(j\omega)\$.
Let's say there is a frequency \$\omega_0\$ and that \$F(j\omega_0)\$ has a very large value. This means that at \$\omega_0\$, \$f(t)\$ and \$\big(\cos(\omega_0 t) - j\sin(\omega_0 t)\big) \$ are very similar to each. This in turn tells you that a sinusoidal component with frequency \$\omega_0\$ is very present in \$f(t)\$. By plotting \$F(j\omega)\$ as a function of \$\omega\$ you see which frequencies are contained in \$f(t)\$ and also how much each frequency component is present. For this reason, the Fourier transform is an excellent tool for frequency analysis of signals.
The same thought process holds for the Laplace transform, however, instead of looking at how similar \$f(t)\$ is to sinusoids, now we look at how similar \$f(t)\$ is to exponentially decaying/growing sinusoids. So \$F(s)\$ is a function telling you how similar \$f(t)\$ is to \$e^{-st}\$ for any \$s\$ and doesn't have the same intuitive usefulness as the Fourier transform does for signal analysis.
However, the Laplace transform is very useful for analysing systems since it allows you to use the initial conditions at \$t=0^-\$ for finding the response of a system. The system response you obtain also contain transient components, and not just the steady state response.
The Fourier transform lacks both of these things for system analysis.
Conclusion
\$F(j\omega)\$ tells you how similar a function \$f(t)\$ is to sinusoids. The Fourier transform is excellent for signal analysis.
\$F(s)\$ tells you how similar a function \$f(t)\$ is to exponentially decaying/growing sinusoids. The Laplace transform is excellent for system analysis.
