Why are some group representations much easier to compute discrete logarithm for? The multiplicative group mod $p$ is isometric to the additive group mod $p-1$, yet computing discrete logarithms in the additive group is easy and completing discrete logarithms in the multiplicative group is usually hard.

What is it about group representation that makes group representation matter so much for the complexity of computing discrete logarithms?
Well, the group being isomorphic doesn't imply that the isomorphism is efficiently computable. If $G \simeq H$ via $\phi : G \rightarrow H$ and $\phi$ is computable, then indeed, DLOG is no harder in $G$ than in $H$ because you can transform the instance.

In the case you mentioned, we have $G = \mathbb Z/(p-1) \mathbb Z$ and $H = (\mathbb Z/p\mathbb Z)^\times$. A efficient morphism $\phi: G \rightarrow H$ is known, it called `fast exponentiation'. So DLOG is no harder in $G$ than it is in $H$. But computing a reversed morphism $\psi: H \rightarrow G$ seems harder...