As described e.g. in this post, quantum gate teleportation can be framed as a variation of quantum state teleportation where a gate is applied beforehand on the receiver, and this results in the final state being teleported being slightly different.
More precisely, if $|\psi\rangle$ is the state to send, $|\Psi\rangle\equiv\sum_i |i,i\rangle$ the shared maximally entangled state, then the standard state teleportation protocol can be seen as consequence of the fact that we can write (ignoring normalisation constants): $$|\psi\rangle|\Psi\rangle = \sum_a |\Psi_a\rangle\otimes (U_a^\dagger|\psi\rangle), \qquad |\Psi_a\rangle\equiv (U_a\otimes I)|\Psi\rangle,$$ where is a(ny) basis of unitary matrices, so that $\operatorname{Tr}(U_a^\dagger U_b)=d\delta_{ab}$, with $d$ dimension of the underlying space. This formulation is nicely described e.g. in this answer.
Similarly, for gate teleportation, we can write $$|\psi\rangle(I\otimes U)|\Psi\rangle = \sum_a |\Psi_a\rangle\otimes (UU_a^\dagger|\psi\rangle),$$ and we thus see how the "gate to teleport" $U$ is applied to the (corrected) teleported state. One can now go ahead and correct the teleported state as usual. Observing that $UU_a^\dagger|\psi\rangle=(UU_a^\dagger U^\dagger) U|\psi\rangle$, we see that to obtain $U|\psi\rangle$ we need to implement a correction of the form $UU_a^\dagger U^\dagger$ for some $a$.
The question is: if we assume to be able to implement local operations on the register containing $|\psi\rangle$, why not just apply directly $U$, either before or after performing standard state teleportation? After all, we are assuming to be able to apply $U$ to (a part of) $|\Psi\rangle$, so why not apply it to $|\psi\rangle$ itself? What's the context where this makes sense?
