Define: $\operatorname {wo}^n(x) \iff \forall y (y \in^n x \to \operatorname {wo} (y))$
Where $\operatorname {wo}(y)$ refers to $y$ being well orderable.
Where $y \in^0 x \iff y=x \\ y \in^{n+1} x \iff \exists z (z \in^n x \land y \in z)$
n-well ordered choice: for $n=0,1,2,...$, for every set $x$ of nonempty sets, if $\operatorname{wo}^n(x)$, then $x$ admits a choice function on it.
If we add this schema to axioms of $\sf ZF$, would it entail axiom of choice?
$2$-well ordered choice is enough to imply AC.
Let $α$ be any ordinal, and look at $\mathcal P^2(α)\setminus\{\emptyset\}$.
We have $x\in^2 \mathcal P^2(α)\setminus\{\emptyset\}⇒∃z⊆\mathcal P(α)\;(x\in z)⇒x\subseteq α⇒x\text{ is well orderable}$.
A choice function on $\mathcal P^2(α)\setminus\{\emptyset\}$ induce a well ordering on $\mathcal P(α)$.
"powerset of well-orderable set is well-orderable" is famously equivalent to the axiom of choice.
No. In the Cohen model every family of well orderable sets admits a choice function. So $1$-well ordered choice holds, but countable choice fails.
