I'm going to assume the following:
Shared vertices are allowed.
The division is into pairs of triangles, not ordered pairs, so that $(ABC, DEF)$ and $(DEF, ABC)$ are "the same". Furthermore, order of verts in a triangle doesn't matter, so that these are both the same as $(BAC, DEF)$.
With that in mind, let's let $K(n)$ be the number of ways to form a single triangle from a polygon with $n$ vertices. You should, I hope, be able to work this out, and let $S(n)$ be the number of ways of forming two nonintersecting triangles.
Each triangle (like the right-hand one in your picture), when excised from the polygon, leaves behind three smaller polygons, although in some cases (like the left-hand one in your picture), these smaller polygons are vacuous. The "other" triangle of the pair must lie in one of these three remainders, so you get a recurrence: for a fixed "first" triangle, there are $K(a) + K(b) + K(c)$ possible second triangless, where $a$, $b$, and $c$ are the sizes of the three "remainder" polygons. You need to sum these over all possible "first" triangles, and then divide your result by two to account for the arbitrariness of calling one triangle the "first" one.
That's a wicked ugly recurrence, but it's probably not hopeless to at least get a big-O bound in the numbers.