Face-walks in rotation systems for graphs
If the sequence $v_0e_1v_1...$ is extended indefinetely without going back to $v_0e_1$, then since the graph is finite, some couple vertex-edge is repeated and hence it is periodic (since the facewalk always continues the same way after meeting this couple vertex-edge), so it looks like $v_0e_1v_1e_2.. v_{i-1}e_{i}v_ie_{i+1}...v_{i+j}e_{i+j+1}v_ie_{i+1}...v_i...v_i$ and so on indefinitely. We can assume that $e_i\neq e_{i+j+1}$ otherwise the loop never starts. 
But then we have $e_{i+1}=\pi_{v_i}(e_i)$ and $e_{i+1}=\pi_{v_i}(e_{i+j+1})$, so $e_i=e_{i+j+1}$, yielding a contradiction. (I intentionnaly left out the $\lambda$ for the sake of clarity, but it works the same for non-pure rotation systems by replacing couple (vertex-edge) by triple (vertex-edge-value of $r_i$ when arriving at the vertex)).
For an edge $e$ attached to vertices $v_1$ and $v_2$, $e$ appears in the facewalks starting by $v_1 e$ and $v_2 e$ (possibly the same facewalk !). Moreover, it only appears in facewalks including $v_1 e$ and $v_2 e$ (where would it come from otherwise ?), so that makes exactly two facewalks (or twice the same). Edit : To be more precise, a facewalk is uniquely defined by a triple (vertex-edge-value of $r_i$ when arriving at that edge). But arriving at $v_1 e$ with orientation (value of $r_i$) -1 yields the same facewalk (up to reversing it) as arriving at $v_2 e$ with orientation +1, so there really are only two different facewalks passing through $e$.