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]\$.