Edray Goins gave a great colloquium today about his work on dessins d’enfants. And in this talk there was a picture that surprised me. It was one of the ones on the lower right of this poster. Here, I’ll put in a screen shot:
Let me tell you what you’re looking at. You are looking for elliptic curves E admitting a Belyi map f: E -> P^1, which is to say a map ramified only over 0,1, and infinity. For each such map, the blue graph is f^{-1}([0,1]), the preimage of the line segment joining o and 1 in P^1(R).
In four of these cases, the graph is piecewise linear! I didn’t know there were examples like this. Don’t know if this is easy, but: for which Belyi maps (of any genus, not just genus 1) is f^{-1}([0,1]) a union of geodesics?
Quomodocumque 
It seems to me that this happens if an invariant differential on the elliptic curve descends to a real meromorphic differential on P^1\{0,1,infinity}, regular along (0,1). The point is that you want a tangent vector to your path, evaluated at the invariant differential, to have locally constant argument; this is evidently true if your differential is the pullback of a real differential.
Yes, I think you can *make* a singular flat metric which is geodesic by thinking of each triangle as Euclidean- if you make them regular triangles, then you get a sextic differential (since angles are multiples of pi/3 = 2pi/6). You can also ask what kind of abelian differentials you get from this by lifting to the natural degree 6 cover. I think this is a great question!
Jordan: if you draw the dessin in the correct geometry (sphere, Euclidean plane, or hyperbolic plane), a dessin is always geodesic. The whole point is that it is obtained by a (geodesic) triangulation. So I would guess that the two non-geodesic ones are with triangle groups with negative curvature.
Like Jayadev Athreya said, a dessins d’enfant is fundamentally taking a number of equilateral triangles (literally, like Polydrons), assembling together according to the gluing pattern, and then uniformizing the result. This is a lot clearer to see in your pictures if you draw the inverse image of the whole real axis, not just the segment [0,1].
To get these extended pictures (combinatorially), take the bipartite graph that is drawn above, add a dot of a third color (purple) in the center of each face, and connect it to all the red and blue dots around the face.
On three out of the four straight-line diagrams above, you get a tiling of a certain torus by equilateral triangles this way. (The pictures are drawn with a shear to draw the torus inside of a square; you need to undo that for the triangles to appear equilateral.) In these cases uniformization does nothing: the assembly of equilateral triangles is already flat.
The remaining straight-line picture, in the middle of the bottom row, becomes a tiling by 45-45-90 triangles. It isn’t as obvious that uniformization makes that flat, but the Schwarz reflection principle proves it. (For that matter, the Schwarz reflection principle is a way to see that the description by assembling equilateral triangles is correct: take the Riemann mapping from the upper half-plane to an equilateral triangle, and extend by reflection.)