I am talking in the context of quantum mechanics. Is there a case where $\hat{V}$ does not simply multiply the function it operates on with itself, $V(x,t)$?
For the harmonic oscillator: $\hat{V}\psi = \frac{1}{2}m\omega^2x^2\psi$. It just multiplies the wavefunction. Is there any case where $\hat{V} = \partial_x$, for example? Does it even make any physical sense to have a potential energy function depend on the wavefunction?
Not quite what you asked for in terms of momentum-dependent potentials but a function of $x$ such as $V(x)$ acts by multiplication on $\psi(x)$ in position space but it will act by derivative $\hat x\to -i\hbar \partial_p$ on the momentum wavefunctions $\psi(p)$. On such functions it is the momentum that acts by multiplication: $\hat p\psi(p)=p\psi(p)$.
The usual example is the linear potential where $$ \hat H=\frac{\hat{p}^2}{2m}+ \alpha \hat x $$ with the Schrodinger equation $$ \left(\frac{p^2}{2m}-i\alpha\hbar\frac{d}{dp}\right)\psi(p)=E\psi(p) $$ in momentum space.
The solution in position space is an Airy function. In momentum space integration is immediate.
