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Why does the TPS of the digraph $G$ is the sum of the TPS of its subdigraphs?

Let $G=(V,E)$ be a graph. Given two subgraphs $H=(V_1,E_1)$, $K=(V_2,E_2)$ of $G$, then the TPS of $G$ is defined as $PS(G)=PS(H)+PS(K)$
Why does the TPS of the digraph $G$ is the sum of the TPS of its subdigraphs?
It has a beautiful proof, but I couldn’t understand it

A:

By definition, a TPS of $G$ is a distribution function $f$ from $E$ to $\mathbb{R}$ (e.g., a nonnegative real-valued function defined on $E$ such that for every $e\in E$, $f(e)\geq 0$ and $\sum_{e\in E}f(e)=1$).
Given two subgraphs $H=(V_1,E_1)$, $K=(V_2,E_2)$ of $G$, you want to prove the following: if $f_H$ and $f_K$ are TPS of $H$ and $K$ respectively, then the TPS of $G$ is given by
$$
f=f_H\oplus f_K
$$
where $\oplus$ is the binary operation defined for nonnegative reals by
$$
x_1\oplus x_2=x_1+x_2\ \ (\mathrm{mod
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