Push relabel algorithms in flow networks
You just have to reorder the elements of $U\times U$, because the double sum contains each edge $(u,v) \in U\times U$ exactly once:

$$\sum_{u \in U}\sum\limits_{v \in U}f(v, u) =
\sum_{(u,v) \in U\times U}f(v, u) = 
\sum_{(u,v) \in U\times U}f(u, v) = 
\sum_{u \in U}\sum\limits_{v \in U}f(u, v)$$

The fist sum sorts the edges by their first vertex, the third sum (of the question's equation) sorts the edges by their second vertex. This is the simplest imaginable form of a double counting argument.