When to use Fourier, Laplace and Z transforms? If we have an LTI system, with an input signal $x(t)$, impulse response $h(t)$ and output $y(t)$, I was under the assumption that if the input and impulse response were continuous in time, then you would use the FT on $x(t)$ and the Laplace transform on $h(t)$, and multiply them together to find the output $Y(f)$ in the frequency domain.

Likewise, if $x[n]$ was a discrete input signal, $h[n]$ was a discrete impulse response, and $y[n]$ a discrete output, then I though we would use the DFT on $x[n]$, z transform on $h[n]$, then multiply them together to get $Y(k)$ which is a discrete output in the frequency domain. 

However I've seen people saying we would use the Fourier transform on the impulse response to get the transfer functions here so it's kinda thrown me off a little bit. Could anyone clarify this?
It's natural consequence of applying a transform to a convolution relation. The output $y(t)$ of an (continuous-time) LTI system is described by a convolution integral :

$$y(t) = h(t)\star x(t) = \int_{-\infty}^{\infty} x(\tau) h(t-\tau) d\tau $$

And when you apply a Fourier transform on this relation, it turns out to be a multiplication in the transform domain such as:

$$ Y(j\omega) = \mathscr{F}\{ y(t) \} = \mathscr{F}\{ h(t) \star x(t) \} $$
$$ Y(j\omega) = \mathscr{F}\{ h(t) \} \cdot \mathscr{F}\{ x(t) \}  = H(j\omega)\cdot X(j\omega) $$

Similary you can also apply a Laplace transform on it :
$$ Y(s) = \mathscr{L}\{ y(t) \} = \mathscr{L}\{ h(t) \star x(t) \} $$
$$ Y(s) = \mathscr{L}\{ h(t) \} \cdot \mathscr{L}\{ x(t) \}  = H(s)\cdot X(s) $$

And exactly the same happens for discrete-time LTI systems.