On the Usual Orientation of Cubic Graphs in Random Construction of Riemann Surfaces In "Random Construction of Riemann Surfaces", Robert Brooks and Eran Makover say :

Definition 2.1 A left-hand turn path on $(\Gamma, \mathcal O)$ is a closed path on [the cubic graph] $\Gamma$ such that, at each vertex, the path turns left in the orientation $\mathcal O$.
The genus of $S^O(\Gamma,\mathcal O)$ is given by 
  $$
\text{genus}=1+\frac{n-l}2
$$
  [$l$ is the number of left-hand paths.]
Note that the usual orientation on the $3$-regular graph which is the $1$-
  skeleton of the cube contains six left-hand-turn paths, giving that the associated
  surface is a sphere with six punctures, while a choice of a different
  orientation on this graph can have either two, four, or six left-hand-turn
  paths, so that the associated surface can have genus $0$, $1$, or $2$.
  Thus, the
  topology of $S^O(\Gamma,\mathcal O)$ is heavily dependent on the choice of $\mathcal O$.

There are $2^{8}$ orientation. What is the usual one? How do these paths look like. An illustrative example is sought, because my brain starts to feel like a punctured sphere...
EDIT:
Does the usual orientation mean that we just assign a "$+$"-orientation, which should mean that we look at it like it is displayed? A "$-$" would flip the meaning of left and right. In this case the following would give the right number of left-hand cycle:
$\hskip1.5in$
 A: The picture that you have is completely right, I think, by the fact that you got the "usual orientation" induced from the orientation of the ambient $3$-dimensional space. Now you can walk around each face by turning always to the left - you get the $6$ left turning paths that the paper speaks about. To get the idea of what other orientation mean, please keep in mind that the paper defines orientation locally as some chosen cyclic order of edges around each vertex. Since you have $3$ edges, and $3$ cyclic shift of thereof, you have only $3!/3=2$ possible orientations around each vertex, and thus $2^8$ possible choices for the cube. Basically, this is what you say on your own: I'm sure you get everything correctly, just hesitate by some reason. 
A: There is a way of doing so by using permutations associated to vertices and to half edges, but this is a bit too long to describe in detail to be a comment. In short: imagine every vertex has 3 "half-edges" around it. You may think these half edges are numbered anyhow. Then an orientation is a choice of independent 3-cycles, and let S be the product of these cycles. Let us now glue together the half-edges: each glueing gives us a transposition. Let T be the product of these independent transpositions. Then the number of cycles in ST is the number of left-turn paths in the graph (provided T is an imprimitive permutation, as far as I remember, which is necessary for the surface to be connected). I'm not quite sure though if this is the way you want to compute things without drawing ... 
