There are two ways to understand this. Well. Three.

If $X$ is a set, then $\mathcal P(X)$ admits a choice function.^{1} This is equivalent to the statement that $X$ can be well-ordered. This means that all the ordinals are "small", and since in $\sf ZF$ the axiom of choice is equivalent to saying that $\mathcal P^2(\alpha)$ admits a choice function for all ordinals, we know that the failure of choice is just saying that for some $\alpha$, $\mathcal P(\alpha)$ is not small anymore.

If you want this ordinal to be explicit, we can introduce many various axioms that make it so. Most famously, Determinacy implies that $\omega$ already fails that. As do many other axioms.

If $X$ is a set, then we can think of $X$ as small if any family of sets indexed by $X$ admits a choice function. Here we have that finite sets are small; $\sf AC_\omega$ states that countable sets are small; and we can have $\sf AC_X$ for non well-orderable sets $X$ just as well.

Again, we can make this more explicit, as $\sf AC$ fails, by stating that $\sf AC_\omega$ fails, or even that for any infinite set $X$, there is a family indexed by $X$ which does not admit a choice function.

We can improve upon this by also providing limitations on how big the sets *inside the family* are allowed to be, e.g. "every countable family of finite sets admits a choice function", but not so for anything larger: either in index or in content.

We can mix those two. We can think of a set as being small if any well-ordered family of subsets admits a choice function, for example. Or any well-ordered family of subsets by a small ordinal (in the sense of (1), that is) admits a choice function. Etc. etc. etc.

There's a lot of variety here. And it gets weirder and weirder, and more and more explicit as we dive into this rabbit hole.

At this point, you might argue, none of these are "constructive negations of choice". Insofar that none of them really pinpoint the exact failure. This is why more structural axioms, such as $\sf AD$, do offer a modicum of success here. Since $\sf AD$ provides us with a fairly rich theory of how badly things get. If you couple it with $V=L(\Bbb R)$, we can say even more about how terrible things can get.

But we can turn our heads to a different path. For example, Monro proved in

*Monro, G. P.*, **Decomposable cardinals**, Fundam. Math. 80, 101-104 (1973). ZBL0272.02085.

That it is consistent that an infinite set is well-orderable if and only if it cannot be written as the union of *two* sets of strictly smaller cardinality. In the Cohen model, which Monro studies, this is true, and we have a relatively straightforward description of this odd partition.

There are many axioms like that, which infuse the universe of sets with a sense of chaos, either in the structure of "what type of families have a choice function" or "what kind of sets can be well-ordered". It seems that you're not entirely clear as to what you're looking for exactly, but there's a lot of these things in the literature.

I suggest starting with Herrlich's "Axiom of Choice" book to read about "disasters" and go from there.

*Footnotes.*

- Yes, $\varnothing$ is an issue. But we can just agree that a family of sets $A$ admits a choice function if $\prod (A\setminus\{\varnothing\})$ is non-empty.

Effective implications between "finite" choice axioms, pp. 439-458 in Mathias/Rogers (editors), Cambridge Summer School in Mathematical Logic, Lecture Notes in Mathematics #337, Springer-Verlag, 1973 (MR 50 #12725; Zbl 279.02047). $\endgroup$1more comment