Consider the following three parts of the sphere: let $P_A$ be the lune created from the triangle $ABC$ plus the triangle adjacent across the $BC$ line segment, plus the opposite lune (on the opposite side of the sphere), and similarly for parts $P_B$ and $P_C$.
The area of $P_A$ is $4\alpha R^2$: the total area of the sphere is $4\pi R^2$, and the area of $P_A$ is certainly proportional to $\alpha$.
Notice now that $P_A\cup P_B\cup P_C$ is the entire sphere, and that $P$'s intersect at the triangle + the opposite triangle. We thus have:
area($P_A$) + area($P_B$) + area($P_C$) = area of the sphere + 2 area of the triangle +2 area of the opposite triangle.
As the two triangles have the same area, you get your formula.
I'd ignore the radius $R$ (i.e. take the earth's radius as unit of length) and then convert to Cartesian coordinates:
\begin{align*}
x&=\cos\phi\cos\lambda\\
y&=\cos\phi\sin\lambda\\
z&=\sin\phi
\end{align*}
These vectors point from the center of the sphere to the vertices of your polygon. Two such vertices are joined by a greatcircle arc. That greatcircle is the intersection of your sphere with a plane through the origin. That plane through the origin is defined by its normal vector. The cross product of two vectors is perpendicular to both of them, so you can use the cross product to compute these normal vectors. Writing $v_i=(x_i,y_iz_i)$ for the vertices, you get the normals of the edges as
$$n_i = v_i\times v_{i+1}$$
The angle at a vertex is equal to the angle between the corresponding normal vectors. You can compute that using the dot product, dividing by the lengths to ensure normalization:
$$\alpha_i = \pm\arccos-
\frac{n_{i-1}\cdot n_{i}}{\lVert n_{i-1}\rVert\,\lVert n_i\rVert}$$
The minus sign in there is because for a normalized dot product of $1$, both normals point in the same direction, so you have zero change in direction but the inner angle there is $180$.
So which sign should you choose for the $\pm$ in there? For that, look at the determinant of the $3\times3$ matrix formed by $v_{i-1}$, $v_i$ and $v_{i+1}$. The sign of that matrix will tell you whether the triangle these three vectors form on the surface is oriented clockwise or counter-clockwise. Take the sign from this and the value from the $\arccos$ above and you should be almost done.
As usual with signs, you have a decision to make: which sign is which? Well, one choice of sign (e.g. exactly as described above) will lead you one polygon, the other (with the sign of the determinant flipped, or eqivalently the order of vectors inside the determinant reversed) leads to the complementary polygon. One will have all its vertices in clockwise order, the other in counter-clockwise order. Trying this out on a tiny example (e.g. three points forming three right angles) will tell you which one is which.
Best Answer
Yeah, you can always break a spherical n-gon in n−2 spherical triangles... You know, an interesting fact about higher dimensional spherical and euclidean simplices is that not all things generalize. For instance there is no relation between angle sums and volume in spherical simplices in the n-dimensional sphere for odd n (see here), and the fact that the sum of the angles is π for euclidean triangles does not generalize (see here) even in 3 dimensions...