The implication $ACC \implies UCC$ is irreversible in $ZF$. This follows again from the transfer theorem of Pincus:
$UCC$ is an injectively boundable statement, see note 103 in "Consequences of the axiom of choice" by Howard & Rubin. Right after the theorem in pp. 285 and its corollary, there are examples of statements of this kind, one of which is form 31 (which is precisely $UCC$). The fact that this is the case follows in turn from the application of lemma 3.5 in Howard, P.-Solski, J.: "The strength of the $\Delta$-system lemma", Notre Dame J. Formal Logic vol 34, pp. 100-106 - 1993
It is known that $¬ACC$ is boundable, and hence injectively boundable.
Then we can apply the transfer theorem of Pincus to the conjunction $UCC \wedge ¬ACC$ and we are done.
SECOND PROOF: Browsing through the "Consequences..." book I've just found another less direct proof of the same fact. I'm adding it here to avoid the interested person from looking it up for himself (especially since the AC website is not working these days). It involves form 9, known as $W_{\aleph_0}$: a set is finite if and only if it is Dedekind finite. Now, $ACC \implies W_{\aleph_0}$ is provable in $ZF$ (in fact this was already proved by Dedekind), while $UCC$ does not imply $W_{\aleph_0}$. The latter follows from the fact that in the basic Fraenkel model, $\mathcal{N}1$, $UCC$ is valid while $W_{\aleph_0}$ is not, and such a result is transferable by considerations of Pincus that can be found at Pincus, D.: "Zermelo-Fraenkel consistency results by Fraenkel Mostowski methods", J. of Symbolic Logic vol 37, pp. 721-743 - 1972 (doi:10.2307/2272420)
I like this question very much.
I find the difference between the two readings of $\forall x\exists y\ P(x,y)$ to be similar to the issues often brought up by questions of uniformity in mathematical existence assertions. So this may be a good keyword leading to further discussion. For example, the discussion of uniformity in my answer to the question Can a problem be simultaneously NP and undecidable?.
To explain a little, in the context of computability theory, say, in the ordinary mathematical reading of $\forall x\exists y\ P(x,y)$, there is no reason to think that there should be a computable function mapping each $x$ to such a $y$. But if we assert that the existence claim is uniform, then this means something much closer to the semantics about which you inquire, namely, that there should be a computable function allowing us to pass from $x$ to a witness $y$.
In ZFC set theory, the axiom of choice can be interpreted as the assertion that all existence claims of the form $\forall x\in u\exists y\ P(x,y)$ are uniform, so that there is a function $f$ with domain $u$ so that $P(x,f(x))$ for each $y\in u$. In class theories such as KM, the global axiom of choice is the assertion that all assertions of the form $\forall x\exists y\ P(x,y)$ are uniform, so that there is a class function $F$ with $P(x,F(x))$ for all $x$.
Since the issue of the uniformity of existence assertions is often central in many mathematical domains, I propose that this is the answer to your question.
Best Answer
There is about half a chapter devoted to this in Jech The Axiom of Choice, the key point is Theorem 7.15 which gives a condition for $\mathsf{AC}(n)\implies\mathsf{AC}(m)$. You may want to look around there (p.111 for the theorem).