How to derive an orientation from a triangulation? Suppose some connected, closed 2-manifold $M$, and suppose I have a triangulation $t:M\rightarrow S$ where $S$ is a homeomorphic simplicial complex such that for each individual 2-cell I can specify a cyclic order on the vertices so that any 2 triangles sharing an edge disagree on the order of their 2 shared vertices, illustrated.
Does this necessarily imply that $M$ is topologically orientable, in the sense that for each point $x \in M$ there is a local orientation i.e. choice of generator $\mu_x \in H_2 (M |x)$ such that there is an open ball $B$ around $x$ where all points $y \in B$ have local orientations $\mu_y$ that are the images of one generator $\mu_B$ of $H_2(M|B)$ under the natural maps $H_2(M|B) \rightarrow H_2(M|y)$ ?
I feel like I'm missing something very obvious. I understand the geometric intuition as to why a consistent choice of clockwise or counterclockwise would induce opposite orientations on each edge. I can also see how, for any given triangle, an order on its vertices could produce a choice of generator for any point on the interior of the triangle, but I don't understand how to prove consistency across the entire manifold. Specifically, how do I find the orientation of points lying on the edges and vertices of the triangulation? And why doesn't this work if there isn't an ordered triangulation where all the edges disagree?
 A: There is a constructive answer and a nonconstructive one:
The constructive answer is that a local orientation in the 2-dimensional case is the same thing as coherently picking oriented loops around each point. If I take an interior point of a simplex, simply use the simplex itself with the orientation determined by the ordering.
Now if we pick an interior point of an edge, we may construct a loop as follows, composed of 4 segments. The first segment is a perturbation of the edge into the interior of the first simplex. The second segment is the perturbation of the edge into the interior of the second simplex. The third and fourth segments are tiny edges connecting the end points of the first 2 segments. This construction has an ordering given by using the induced ordering of the OTHER simplex than the one it was perturbed into  and extending to the short edges. The fact that this is an orientation of the loop comes from the fact that the edges have opposite induced orientations. Now if you want to show the axioms for an orientation, you should just show that if instead you had picked the orientation given by slightly enlarging the first simplex, you get a homotopic loop and similarly for the second. The point being that this intermediate loop is an inbetween between the orientations determined by enlarging either of the two simplices, and so it is easier to see that all of these choices are homotopic.
At this point, we just have to extend to the vertices which you can think about.
The nonconstructive answer is that picking a cyclic ordering with your properties specifies a nonzero cycle in $C_2(M)$. Namely, take the sum of all the simplices corresponding to your triangulation. The fact that the orientations disagree whenever an edge is shared implies that when I apply the boundary map, every 1 simplex will occur twice with opposite signs, meaning the boundary is $0$. Since there are no 3-simplices, this implies that the $\mathbb{Z}$ top homology is nontrivial which is known not to happen for nonorientable manifolds by careful application of universal coefficients and Poincare duality.
