Is Axiom of Choice equivalent to its version for families of sets, indexed by ordinals? Is Axiom of Choice equivalent to the following statement?

Axiom of Ordinal Choice: For any ordinal $\lambda$ and any indexed family of sets $(X_\alpha)_{\alpha\in\lambda}$ there exists a function $f:\lambda\to\bigcup_{\alpha\in\lambda}X_\alpha$ such that $f(\alpha)\in X_\alpha$ for all $\alpha\in\lambda$?

 A: It is strictly weaker than choice. This is explained in Asaf Karagila's answer at MSE: the $L(\mathbb{R})$ of $L$ + $\aleph_1$-many Cohen generics witnesses this.
(There the principle is phrased for well-ordered index sets, but you can always pass from a well-ordered index set to an ordinal index set.)

As Taras Banakh comments below, there is a superficially-similar fact: namely that over NBG without global choice, global choice is equivalent to $Ord$-indexed choice. The proof is simple: for $\alpha\in Ord$ let $C_\alpha$ be the set of well-orderings of $V_\alpha$. From a choice sequence $(\triangleleft_\alpha)_{\alpha\in Ord}$ we get a well-ordering of $V$ as follows: set $x\prec y$ iff $rk(x)<rk(y)$ or $rk(x)=rk(y)$ and $x\triangleleft_{rk(x)}y$.
Note the key role of set-sized choice above. More abstractly (and being a bit sloppy with details), what's being used here is that in NBG without global choice every class is "locally well-orderable" in the sense of being a well-ordered union of well-orderable sets (via Foundation plus Set-Choice). We deduce global choice from well-ordered-class choice by first applying WOCC to well-order the locally-well-ordered indexing class of our global choice instance, and then applying WOCC to that.
However, in ZF we don't have anything like this. For example, amorphous sets are not locally well-orderable in the relevant sense. Instead, we get the following:

(ZF) Suppose $\lambda$ is a limit ordinal and full choice holds in $V_\alpha$ for each $\alpha<\lambda$. Then the following are equivalent:

*

*Every $s\in (V_\lambda\setminus\{\emptyset\})^\lambda$ has a choice function.


*$V_\lambda$ is well-orderable.

