Shortest path with exactly $k$ edges
Since you allow non-simple paths(i.e. walks), seems like a dynamic programming algorithm will work. 

For each $1 \le m \le k$, and every vertex $u$, we compute $D[m,u]$ where $D[m,u]$ is the weight of the shortest walk of length exactly $m$ starting at $v$ and ending at $u$. We are looking for $D[k,w]$.

This can be computed as

$$D[m+1, u] = \min_{x \in Pred(u)}\{D[m,x] + w[x,u]\}$$

Where $Pred(u)$ (predecessor) is the set of vertices which have an outgoing edge to $u$ and $w[x,u]$ is the weight of edge $x\to u$. You start with $D[1,u] = w(v,u)$ (allowing $\infty$).

You can add auxiliary structures to find the actual walk.

Running time: $O(k |E|)$. 

This is because for each number of required edges from 1 to k, we go through the indegree of each vertex and in total, we visit all $|E|$ edges. We visit all $|E|$ for each number of the required edges ($k$) so the final run time is $O(k |E|)$ .

btw, if we wanted only simple paths, then this is probably $NP$-Hard, as we can set $k=n$ and reduce some variant of Hamiltonian Path problem to it.