"Indexed" version of compactness and Axiom of Choice

Your concern about several indices giving the same set in the cover isn't really a problem. Given an indexed open cover, "plain" compactness provides a subcover involving only finitely many of the open sets but possibly (as you noted) involving infinitely many of the indices. Fortunately, the axiom of choice is not needed in order to choose from finitely many sets. So you can, without using the axiom of choice, choose one of the many indices for each of the finitely many open sets in your subcover.

They're equivalent, and you don't need choice to prove that they're equivalent.

It's fairly clear that 'indexed compactness' implies compactness. Conversely, we can turn any subset $\mathcal{U} \subseteq \mathcal{T}$ into an function $f : I \to \mathcal{T}$ by taking $I = \mathcal{U}$ and taking $f$ to be the inclusion map $\mathcal{U} \hookrightarrow \mathcal{T}$. For this choice of $I$ and $f$, the statement of 'indexed compactness' reduces to the usual statement of compactness.