The axiom of choice is needed when there is no uniform choice from the sets that you wish to choose from.
Given any non-empty set $A$, $2^A$ is a product of the set $\{0,1\}$ over the index set $A$. This product is never empty, we can always insist on choosing $0$.
Another example is whenever $A$ can be well-ordered, we can always choose from families of subsets of $A$. How? Simply fix a well-order of $A$, then every non-empty set has a least element and that is our uniform choice. So uniform choice means that we can more or less formulate what would be the choice from each set (there is some delicacy here still).
So what about $\Bbb{R/Q}$? Well, it turns out that we don't have a uniform choice from these sets, and it sort of make sense in a way. All the equivalence classes are dense subsets of $\Bbb R$, and we don't have too much information on them beyond that. But this is not a proof that we can't make this choice, it's just reasoning.
The proof that we cannot make this sort of choice comes from the fact that there are models of set theory where the axiom of choice fails, and every set of real numbers is Lebesgue measurable. In such model, certainly there is no Vitali set - that is a system of representatives for $\Bbb{R/Q}$ - and therefore there is no choice function from $\Bbb{R/Q}$.
In fact, just to make things weirder, let me give you the extent of which the axiom of choice is needed. It is possible to have a universe of set theory, where there is a countable set of pairs, but we cannot choose from from each pair. This is because we don't have a uniform way of matching them up with $\{0,1\}$.