Axiom of choice and cartesian product

The set of choice functions of domain $\{X_i|i\in I\}$ satisfying $f(X_i)\in X_i$ for all $i\in I$ has cardinality $\prod_i|X_i|=|⨉_iX_i|$, because this set of functions has an obvious bijection with the Cartesian product $⨉_iX_i$, namely by pairing such a function $f$ with the ordered tuple with $i$th element $f(X_i)$. The axiom of choice is usually formulated as the claim that this set of choice functions is nonempty, but what you've read is that it can be equivalently stated as the claim that the equally large Cartesian product (itself a set of ordered tuples) is nonempty. Well, of course these statements are equivalent, as the sets are equinumerous.

We'd need further information to ensure there are multiple choice functions, because if each $X_i$ is a singleton there won't be. But our switch to focusing on Cartesian products doesn't cause any tuples or functions to "go missing".

When you talk about functions from $I$ to $\cup X_i$, you presumably mean the ones satisfying $f(i)\in X_i$ for all $i\in I$. As I noted above, the "existing" functions are the ones whose associated tuples "exist". How many of these there are depend on the model of ZF. (For example, a model of $ZF\neg C$ implies the existence of $X_i$ for which zero of either exist.)

It sounds like you want to prove the following statement:

• Given a partial function $f:I\to\cup X_i$, there exists a (total) function $F:I\to\cup X_i$ that extends $f$.

Can you see how to define such an $F$, given the existence of some function $C:I\to\cup X_i$?