Does PPAD really capture the notion of finding another unbalanced vertex?
This is an interesting question, and I can only give a partial answer.

It is easy to see that the construction on p. 505 of Papadimitriou’s paper shows the equivalence of AUV with its special case


  MANY ENDS OF THE LINE (MEOL): Given a directed graph $G$ with in-degree and out-degree at most $1$ (represented by circuits as above), and a nonempty set $X$ of sources of $G$, find a sink or a source $v\notin X$.


On the one hand, I find it hard to imagine a transformation of such graphs that could reduce a larger number of sources to one.

However, on the other hand, MEOL belongs to all commonly studied classes containing PPAD except possibly PPAD itself:

First, obviously, 


  MEOL is in PPADS.


I will sketch below an argument that 


  MEOL is in PPA


by reduction to the standard PPA-complete problem (the undirected version of AEOL). Assume we are given $G=(V,E)$ and $X$ as in the definition of MEOL.

If $|X|$ is odd, we can just make the graph undirected, include a matching on all but one vertex from $X$ (as in the argument on p. 505), and pass the result along with the remaining source from $X$ to the PPA oracle.

In general, let $s=|X|$, and $2^k$ be the largest power of $2$ that divides $s$. We define a new graph $G'=(V',E')$ whose vertices are $2^k$-element subsets of $V$. If $A,B\in V'$ are such sets, we put the edge $(A,B)$ in $E'$ if we can enumerate the sets as $A=\{a_0,\dots,a_{2^k-1}\}$, $B=\{b_0,\dots,b_{2^k-1}\}$ in such a way that $(a_i,b_i)\in E$ for each $i&lt;2^k$.

Clearly, $G'$ is a directed graph with in-degree and out-degree at most $1$. An $A\in V'$ is a source (sink) iff it contains a source (sink, resp.) of $G$. (That is, if it contains both, it is an isolated vertex.) So, any such vertex would lead to a solution of the MEOL instance, unless $A$ is a “known source”: that is, $A\cap X\ne\varnothing$. We intend to make the graph undirected, and manipulate it so that the number of known sources is reduced to $1$ by including a matching on the remaining ones.

So, if $A$ is a known source, let $t=|A\cap X|$, which satisfies $0&lt;t\le2^k$. If $t=2^k=|A|$, then simply $A\subseteq X$. The number of such sets is $\binom{s}{2^k}$. Recall that the multiplicity of a prime $p$ in $\binom ab$ equals the number of carries in the addition $b+(a-b)=a$ performed in base $p$. By the choice of $k$, it follows that $\binom{s}{2^k}$ is odd. Moreover, there are polynomial-time bijections between $[0,\binom ab)$, and $b$-element subsets of $[0,a)$. Using this, we can define a polynomial-time matching on all but one of $2^k$-element subsets of $X$. We include it in the graph, which reduces the number of known sources with $t=2^k$ to $1$.

For $0&lt;t&lt;2^k$, the carry-counting formula shows that $\binom st$ is even. Again, we can find an explicit matching on $t$-element subsets of $X$. We extend it to known sources $A$ with $|A\cap X|=t$ by applying the matching to $A\cap X$, and leaving $A\smallsetminus X$ fixed.

In this way, we produce an undirected graph with one known leaf vertex. We ask the PPA oracle for another leaf, and by construction, we can extract from it an answer for the MEOL instance.



As briefly mentioned by Papadimitriou, we can generalize PPA to classes PPA-$p$ for any prime $p$. An example of a PPA-$p$ complete problem is


  AUV-$p$: Given a directed graph $G$, and a vertex whose degree balance is ${}\not\equiv0\pmod p$, find another such vertex.


(See this answer for the equivalence of AUV-$p$ with Papadimitriou’s definition of PPA-$p$.) 

PPA-$2$ is just PPA. The classes PPA-$p$ are assumed to be pairwise incomparable, and incomparable with PPADS. They all include PPAD.

There was nothing particularly special about $p=2$ in the argument I outlined above, and it can be easily modified to yield


  MEOL is in PPA-$p$ for every prime $p$.