In every reference I've seen, the area of an $n$-sided spherical polygon is given by the sum of all interior angles, minus some amount of spherical excess, or
$$\text{Area}(\text{polygon}) = \sum_{i} A_i - (n-2)\pi.$$
This is often derived as repeated application of Girard's Theorem to each of the $n-2$ triangles that compose the polygon, where the total sum of angles over each triangle is equal to the sum of the interior angles of the polygon, and you lose $\pi$ area from the spherical excess of each. This response has a good explanation of the concept, and I think is pretty representative of the discourse.
However, these explanations seem to implicitly assume that the polygon in question is convex, which of course isn't always the case (see this question, which purports to have a counterexample to Girard's Theorem for a nonconvex shape). In particular, I'm interested in the more robust case that comes up in computational geometry, where the polygon may not just be nonconvex, but potentially self-intersections as well. In this case, we presumably have to consider the signed area of each (now possibly overlapping) triangle. The simplest case of this I can think of is an oriented "hourglass" shape, basically a regular spherical square with two vertices swapped, labeled as such. For the algebra that follows, let the shared angle of the spherical square be $\alpha$.
As far as I can figure, the signed area of this spherical polygon should be zero, as the top and bottom "bulb" have the same area and opposite orientation. This intuition is upheld if you triangulate the hourglass, with $\text{SignedArea}(\Delta 123) = \alpha/2 + \alpha + \alpha/2 - \pi$ and $\text{SignedArea}(\Delta 134) = -(\alpha/2 + \alpha/2 + \alpha - \pi)$, since this latter triangle is oriented clockwise on the surface.
However, Girard's Theorem would tell you that $\text{Area}(\square 1234) = 4(\alpha/2) - 2\pi$. In this case you get a value that is equal to the unsigned area of the shape, but I've looked at other nonsimple shapes for which this isn't the case (such as a pentagram formed by taking alternating vertices of a pentagon). Is there any way to account for this changing orientation within Girard's formula? Or is the only option for computing this area to revert back to the original triangulation?