So let's have a graph $A$ with only two vertices $u$ and $v$, and only one edge $e=(u,v)$ between them so $A$ is simple. By Euler's formula, $A$ has only one face and its length $l=2e(G)=2$. But $A$ has only one edge as we assumed, and isn't the path $u$-$v$ enough to bound it?
Attached picture has a slightly more complicated example: 
As I wrote, $G$ has two faces: one is the triangle $v_2v_3v_4$, and the other is the outer face of length $5$. Similar to the one-edge graph $A$ above, I can't see why the $4$-edge walk, $v_{1}v_{2}v_{3}v_{4}v_{2}$, is not enough to bound the outer face.
I am afraid that I might have misunderstood the definition of a face, but I can't see how. Thank you for your help and I appreciate any insight.
I ussualy work with degree of faces, rather than length. But the issue is the same. See this question for exemple.
The edge $v_1v_2$ must be accounted for 2. In general, you should see each edge as having two "sides". If the two are required to bound the face, you should double count the edge.