I've played several games of this and found that, in my random assignments, Alice tends to win. While I haven't proven that this is the winning strategy, I have found a strategy that I have yet to find a way to beat as player 2.
The key insight:
The way I looked at it, you want your moves to have the maximum effect. You'd never want to play a move that is the third corner of triangles that you have already claimed. So the strategy should minimize that. The strategy posted by @Plop does seem to do that, but more that Alice will never make more mistakes than Bob. So I figure you should choose the spot that maximizes the number of triangles that you are "claiming" ie placing your color on one of the corners.
Assuming that from top to bottom and left to right, the playable vertices are A - O. Thus, E, H, and I are the best with 7 triangles. Then D, F, and M are next with 6. After that the 1/3 edges, B, C, G, H, L, and N with 4 triangles. And lastly the corners A, K, and O. Alice should maximize the number of triangles that she claims. So pick the vertex that is in the highest tier and the most triangles that 1) are the second of A's color, 2) are placing a new claim in a triangle where all 3 vertices are open, 3) that have just 1 of Bob's vertices, and 4) have the fewest triangles that have 2 vertices claimed by Bob.
This seems to maximize Alice's wins. Following that strategy, regardless of what Bob does, at least in all of my test games, player 1 will win.