A graph is a typical data structure which is used to represent some relations between nodes. From any graph, we can imagine a flow of a graph $G$. Formally, flow of the graph is defined like follow. For a given graph $G= (V, E)$, flow $f$ is a mapping function $V \times V \rightarrow \mathcal{R}$. To be a flow, there are some requirements. If there is a capacity $c$ for edeges $c : V \times V \rightarrow \mathcal{R}$, which $(u,v) \in E$ iff $c(u,v)$ exists. Then flow of graph can’t exceed that capacity. Which means $f(u,v) \le c(u, v)$. Also it requires two variables which are source $s$ and destination $t$. Then $\sum\limits_{(i, v) \in E, v \neq s,t}f(i, v) = \sum\limits_{(v, j) \in E, v \neq s,t}f(v, j)$. Which means that every value goes in $v$ should go out from $v$ either. Then we can evaluate this flow $f$ as $\sum\limits_{(s, v) \in E}f(s, v) - \sum\limits_{(v, s) \in E}f(v, s)$ $=$ $\sum\limits_{(v, t) \in E}f(v, t) - \sum\limits_{(t, v) \in E}f(t, v)$.
Now, let’s think about a directed graph $G = (V,E)$ and some flow $f$ of $G$. Then we can decompose $f$ to some simple paths $P_1, \cdots, P_N$ and some cycles $C_1, \cdots, C_M$. Which $N + M \le \left\vert E \right\vert$.
Without loosing much of generality, we can assume that $f$’s value is positive or negative. In the proof of flow decomposition, it doesn’t necessarily to be positive or negative. However, let’s assume it to be positive for easy.
Then, if $f$ has a cycle $C$, we can eliminate that $C$ from $f$ without changing value of the flow because it is a cycle. If we reduce every value of edges with smallest value inside of the $C$ then we can remove that cycle. Notice that we can reduce value of edges because we picked the smallest value inside of the $C$, cycle should be disappear after reducing values because one of edge in $C$ should be 0 after reducing values.
If we remove every cycle, it should have some paths because it has a flow. If we remove any flow that has the smallest value, value of flow $f$ should be 0 or some positive value. If it is the first case (flow of $f$ become 0) then it is decomposed. Otherwise, we can do this process again. Therefore, there sould be at least some cycles and paths.
Notice that we should remove at least 1 edge to remove a path or a cycle because it choosed a smallest value on graph. As a result, we can do this at most $\left\vert E \right\vert$.