In the quantum hamiltonian, why does kinetic energy turn into an operator while potential doesn't?

Actually the potential is also an operator. It just so happens that, in the position representation, $\hat x\psi(x)=x\psi(x)$, so that the potential energy operator $V(\hat x)$ acts by multiplication: $V(\hat x)\psi(x)=V(x)\psi(x)$.

Even though it acts “by simple multiplication”, it doesn’t commute with the momentum, a sign that it is still a legitimate operator.

In the momentum representation $\hat x$ acts by differentiation so in this case the potential energy operator becomes a (usually quite complicated) differential operator since one needs to use the formal expansion of potential to convert it to a polynomial.

OP's hunch is exactly right: In QM they are all operators. However each operator may become a multiplication operator in certain representations. E.g. position operators become multiplication operators in the Schrödinger position representation, while momentum operators become multiplication operators in the momentum representation, and so forth.