Set $A_k = \{\gamma \ | \ \omega \cdot k \leq \gamma < \omega \cdot (k+1) \}$, and let $B$ be a set of ordinals such that $B = \{ \alpha_n \ | \ n \in \omega \}$ where each $\alpha_n$ is an at most countable ordinal. Can we say that a set, $\{g_{\alpha_k} \ | \ k \in \omega \}$, of functions exists, such that each $g_{\alpha_k}$ is a surjection from $A_k$ to $\alpha_k$, without any form of choice? Obviously if these were some arbitrary sets then we would need choice, but since they are ordinals I am not sure if choice is necessary or not.
I needed this fact in one of my assignments in which there was no restriction on the usage of choice, so I used it freely. But now I wonder if it can be done without choice, thank you.
No.
If $\omega_1$ is singular, then its cofinality is countable. And this is indeed consistent with $\sf ZF$. Then there is a countable set of countable ordinals, $\{\alpha_n\mid n<\omega\}$ whose supremum is $\omega_1$.
In that situation it follows that there is no way to uniformly enumerate all the $\alpha_n$s. If there would, then their union would be countable.
What is true in $\sf ZF$ is that if $\sup B<\omega_1$, then this is possible. So for the general fact to be true for all countable sets of countable ordinals we need to assume that $\omega_1$ is regular. And for that we need just a bit of choice.