Let denote the graph by $G$ , the number assigned to the edge $e\in E(G)$ - by $n(e)$ and the number that will be assigned to a vertex $v\in V(G)$ - by $n(v)$. Assume that $G$ is a connected graph (if not, we will apply the arguments below to each connected component of $G$). Let $T$ ia a spanning tree of $G$ with a root $v_0$. Assign to $v_0$ , for now being, an arbitrary number $n_0$ , i.e $n(v_0)=n_0$. We will choose $n_0$ later. We can consecutively assign numbers to the vertices of $T$, so that for every edge $xy$ of $T$ holds: $n(xy)=n(x)+n(y)$. Our aim is to prove that the same holds for every edge $e\in E(G)\setminus E(T)$. Let $xy$ is an edge of $E(G)\setminus E(T)$ connecting $x,y\in T$. Consider the closed path $p = x T v_0 T y x$, where $xTy$ means the unique path in $T$ connecting the vertices $x$ and $y$. Of course $p$ may have repeated edges. If the length of $p$ is even then applying the given property an easy calculation yields $n(uv)=n(u)+n(v)$. But the length of $p$ may be odd. So if all these paths are with even length we are done. Now assume $x_0y_0 \in E(G)\setminus E(T)$ so that the path $p = x_0 T v_0 T y_0 x_0$ has an odd length. It's time now to determine $n_0=n(v_0)$ so that to hold: $n(x_0y_0)=n(x_0)+n(y_0)$. An easy calculation shows that it is possible. Now if $xy$ is an arbitrary edge of $G\setminus T$ and the length of path: $p=x T v_0 T y x$ is odd we consider a new closed path: $p'=v_0 T x y T v_0 T x_0 y_0 T v_0$. The length of $p'$ is even as a sum of two paths with odd lengths. For each edge of $p'$ except $xy$ , the number assigned to that edge is sum of numbers assigned to its incident vertices. Now using the given condition an easy calculation shows that it is also true for $xy$. Comment: I think an easy one for a TST. There is a clear motivation and more efforts are needed to write it than to solve it.

Can someone explain how it is easy to conclude $n(uv)=n(u)+n(v)$ if $p$ is even. I think $u$ and $v$ should be $x$ and $y$ as well.