Does choosing a counting function require the Axiom of Choice?

No, you do not need the axiom of choice there. The point is that $X$ is countable iff it has a counting function. You are only making one choice - you are choosing an element from $\{$counting functions on $X\}$, which you know (by hypothesis) to be non-empty.

No, you are wrong. If $X$ is finite, the fact that there is an enumeration of $X$ as $\{x_1,\ldots,x_n\}$ is not due to the axiom of choice for finite sets being a theorem. It is by the definition of finiteness.

Similarly, the definition of $X$ being a countable set, means there is an injection from $X$ into $\Bbb N$. Using existential instantiation we get an enumerating function $f\colon X\to\Bbb N$.