The axiom of choice allows us to make many choices at the same time.
It is true, without the axiom of choice, that if $(A_i,f_i)_{i\in\Bbb N}$ is a sequence of countable sets, and $f_i\colon A_i\to\Bbb N$ is an injection, then $\bigcup A_i$ is countable. This is because we are given the injections into $\Bbb N$.
However in the general case, one has to choose such injection (often a bijection). These injections are not at all definable that we can just write down some (perhaps very complex) formula witnessing their existence, and indeed without the axiom of choice it is possible that there is a family of countable (or even finite) sets whose union is not countable -- exactly because we cannot choose the injections in a uniform way.
On the other hand, when we take the product $A\times B$ we only have to make two choices, $f\colon A\to\Bbb N$ and $g\colon B\to\Bbb N$ - two injections - which is provably possible without the axiom of choice. And we already know how to find a bijection between $\Bbb N$ and $\Bbb{N\times N}$ in a very definable way, so that is enough.
The main issue here is that when we take products, or even unions, of infinitely many sets then we often need some uniformity in the way we order these sets (be it well-order, or otherwise). The axiom of choice allows to prove that such uniformity exists, because we can choose an ordering for each set; but without the axiom of choice it is sometimes the case that we may end up in a universe where such uniformity does not exist, i.e. without some form of the axiom of choice (or a weaker form) we cannot prove a particular statement.