Relation and difference between Fourier, Laplace and Z transforms
The Laplace and Fourier transforms are continuous (integral) transforms of continuous functions.
The Laplace transform maps a function \$f(t)\$ to a function \$F(s)\$ of the complex variable s, where \$s = \sigma + j\omega\$.
Since the derivative \$\dot f(t) = \frac{df(t)}{dt} \$ maps to \$sF(s)\$, the Laplace transform of a linear differential equation is an algebraic equation. Thus, the Laplace transform is useful for, among other things, solving linear differential equations.
If we set the real part of the complex variable s to zero, \$ \sigma = 0\$, the result is the Fourier transform \$F(j\omega)\$ which is essentially the frequency domain representation of \$f(t)\$ (note that this is true only if for that value of \$ \sigma\$ the formula to obtain the Laplace transform of \$f(t)\$ exists, i.e., it does not go to infinity).
The Z transform is essentially a discrete version of the Laplace transform and, thus, can be useful in solving difference equations, the discrete version of differential equations. The Z transform maps a sequence \$f[n]\$ to a continuous function \$F(z)\$ of the complex variable \$z = re^{j\Omega}\$.
If we set the magnitude of z to unity, \$r = 1\$, the result is the Discrete Time Fourier Transform (DTFT) \$ F(j\Omega)\$ which is essentially the frequency domain representation of \$f[n]\$.
Laplace transforms may be considered to be a super-set for CTFT (Continuous-Time Fourier Transforms). You see, on a ROC (Region of Convergence) if the roots of the transfer function lie on the imaginary axis, i.e. for s=σ+jω, σ = 0, as mentioned in previous comments, the problem of Laplace transforms gets reduced to Continuous Time Fourier Transform. To rewind back a little, it would be good to know why Laplace transforms evolved in the first place when we had Fourier Transforms. You see, convergence of the function (signal) is a compulsory condition for a Fourier Transform to exist (absolutely summable), but there are also signals in the physical world where it is not possible to have such convergent signals. But, since analysing them is necessary, we make them converge, by multiplying a monotonously decreasing exponential e^σ to it, which makes them converge by its very nature. This new σ+jω is given a new name 's', which we often substitute as 'jω' for sinusoidal signals response of causal LTI (Linear Time-Invariant) systems. In the s-plane, if the ROC of a Laplace transform covers the imaginary axis, then it's Fourier Transform will always exist, since the signal will converge. It is these signals on the imaginary axis which comprise of periodic signals e^jω = cos ωt + j sin ωt (By Euler's).
Much in the same way, z-transform is an extension to DTFT (Discrete-Time Fourier Transforms) to, first, make them converge, second, to make our lives a lot easier. It's easy to deal with a z than with a e^jω (setting r, radius of circle ROC as untiy).
Also, you are more likely to use a Fourier Transform than Laplace for signals which are non-causal, because Laplace transforms make lives much easier when used as Unilateral (One sided) transforms. You could use them on both sides too, the result will work out to be the same with some mathematical variation.
Fourier transforms are for converting/representing a time-varying function in the frequency domain.
A laplace transform are for converting/representing a time-varying function in the "integral domain"
Z-transforms are very similar to laplace but are discrete time-interval conversions, closer for digital implementations.
They all appear the same because the methods used to convert are very similar.