Suppose $G=(V,E)$ be a directed graph , and let $u,v,w$ be distinct vertices. Suppose there are $k$ edge disjoint paths from $u$ to $v$ in $G$, and $k$ edge disjoint paths from $v$ to $w$ in $G$. The paths from $u$ to $v$ van share edges with the paths from $v$ to $w$. Then, how to show that there are $k$ edge disjoint paths from $u$ to $w$ in $G$ using Menger's theorem or MaxFlow-MinCut.
Update: I have already proved the following statement: Given an integer $k>0$, $G$ has $k$ edge disjoint paths from $s$ to $t$ if and only if there is an $s,t$ flow of value $k$ in $G$.
By the above statement, we know that $G$ has a $u,v$ flow of value k, and $G$ has a $v,w$ flow of value k, but how can we show there is a $u,w$ flow of value $k$ in $G$? This means I have to prove the transitivity of a flow, my thought is to use the flow conservation property of internal vertex to prove that, the internal vertex which joins the $u,v$ flow and $v,w$ flow is $v$. Is this the right approach or there is a better way to approach this?
First, let's give an answer using the MaxFlow-MinCut theorem where we assign each edge a capacity of $1$.
The key observation you already made was: Let $s,t$ be arbitrary nodes, $k\in\mathbb N$. Then there are $k$ edge-disjoint $s$-$t$-paths if and only if there is an $s$-$t$-flow of value $k$.
Assume, we have less than $k$ edge-disjoint $u$-$w$ paths in $G$. Then a maximum $u$-$w$-flow must have a value less than $k$. By the MaxFlow-MinCut theorem there is a $u$-$w$-cut $C$ with capacity less than $k$. Now we do a case distinction on $v\in C$:
Now for an answer using Menger's theorem on edge-connectivity. This states for nodes $x$ and $y$, that the minimum number of edges whose removal disconnects x from y is equal to the maximum number of edge-disjoint paths from x to y.
Assume, there are at most $k'<k$ edge-disjoint $u$-$w$-paths in $G$. Then by Menger's theorem there are edges $e_1,\dots, e_{k'}$ whose removal disconnect $u$ from $w$. As $k' < k$, this does not disconnect $u$ from $v$; nor does it disconnect $v$ from $w$. But then there is a path from $u$ to $w$. Contradiction.