Working in  "[MK][1]-Regularity-Limitation of Size + Replacement for sets", call it the Base theory, let's coin the following axiom:

**Axiom of Super-Choice:**$$\forall \ relation \ R \ \exists F \subset R \ (F: dom(R) \to rng(R))$$ Where: $$R \text { is a relation } \equiv_{df} \forall r \in R \ \exists x,y \ (r=\langle x,y \rangle)\\dom(R)=\{x| \exists y (\langle x,y \rangle \in R)\} \\rng(R)=\{y|\exists x (\langle x,y \rangle \in R)\}$$

  I think this would be equivalent to Global Choice over the Base theory. 

Now if we weaken the above to:

**Axiom of Proper Class Choice:**$$\forall \ relation \ R \ \text{ with proper class rows } \\\exists F \subset R \ (F: dom(R) \to rng(R))$$  

where: $$ z \text{ is a row of }R \iff 
\\ \exists x \in dom(R)[z=\{y| \langle x,y \rangle \in R\}]$$

A relation $R$ is with proper class rows, if  every row of it is a proper class.

>Would that still be equivalent to Global Choice over the Base theory?

>If not, then if we also add Choice over sets [i.e., every set has a choice function], would that result in proving Global Choice?

  [1]: https://en.wikipedia.org/wiki/Morse%E2%80%93Kelley_set_theory