Can a flow be decomposed in a given number of paths?
This is the proof given here, to show that the flow obtained after two iterations of Ford-Fulkerson algorithm can be decomposed into at most $3$ paths and a circulation. It is assumed that at each iteration, the algorithm picks the path that allows maximum flow.

Let the path obtained in first iteration be $P_1$ with flow $f_1$. And the path in the residual graph w.r.t this flow, picked by the algorithm in second iteration be $P_2$ with flow $f_2$. Clearly $f_1 \geq f_2$. Now consider the path $P_1$ with flow $f_1-f_2$. Removing this flow from the total flow, the remaining flow can be described as follows (for each edge $e$).


If $e \notin P_1$ and $e \notin P_2$ then $f(e)=0$.
If $e \in P_1$ and $e \notin P_2$ then $f(e)=f_2$.
If $e \notin P_1$ and $e \in P_2$ then $f(e)=f_2$.
If $e \in P_1$, $e \in P_2$ and $e$ is a forward edge in $P_1$, then $f(e)=2f_2$.
If $e \in P_1$, $e \in P_2$ and $e$ is a backward edge in $P_1$, then $f(e)=0$.


So the remaining flow is now $f_2$-integral, with a total flow of $2f_2$ from $s$ to $t$. We can show in general that if a flow is $r$-integral with a total flow of $kr$ from $s$ to $t$, then this flow can be decomposed into at most $k$ $s$-$t$ paths and a circulation. We can just start from $s$ and move along any edge carrying some flow and reach $t$. This gives an $s$-$t$ path with flow $r$ (or some multiple of $r$). Remove this flow and repeat this process again (walk from $s$ to $t$ along flow carrying edges). Finally, when there is no net flow let from $s$ to $t$, the remaining flow is a circulation. That this process ends in $k$ steps can be shown formally using induction.