For the purposes of this question, a spherical polygon in $\mathbb S^2$ with isometric side pairings is called dihedral if:
- It is star-shaped with respect to some interior point;
- It is homeomorphic to $D^2$ as a polygon, and to $S^2$ after the side pairings are performed;
- Its vertices lie either at the poles or the equator; and
- Under the side pairings, polar vertices are never glued to equatorial vertices, and the sum of the interior angles of a maximal set of equatorial vertices that get glued is either $\pi$ or $2\pi$. (Note that vertices are allowed to be degenerate—that is, to have interior angle $\pi$.)
The simplest two distinct examples of dihedral spherical polygons are bigons with their opposite edges paired, or bigons with two degenerate equatorial vertices added and the edge pairings induced by a reflection in the equator. These two examples are shown below.
I believe that, essentially, these are the only two examples—that is, given any dihedral spherical polygon, one can cut and paste it, consistently relabelling the edges, in such as way as to simplify it to one of the two cases above. Another way of saying this is (I believe) that the quotient of the polygon by the side pairings is isometric to one of the above. Does anyone have an idea of how to prove this, or have a counterexample? In case it helps, an easy example of a simplification is shown below.
First assume that both poles are vertices of the polygon. Then we are up to symmetry in one of the following four cases:
Using suggestive labels (i.e., $n$ for north pole etc.), the perimeter is $nw_1\cdots w_mse_1\cdots e_kn$ for some non-negative integers $m,k$ where $m+k$ is even. In the first figure, we may have $m=k=0$, in which case it corresponds to your first example and we are done. In the second figure, we cannot have $k=0$ as this would not allow us to glue edge $sn$ to anything. Thus we we may now assume $m\ge1$ and $k\ge 1$.
If $nw_1$ is glued to $se_1$ (and $e_kn$ to $w_ms$), the final result is not homeomorphic to $S^2$.
Assume $nw_1$ is glued to $ne_k$ (and $se_1$ to $sw_m$). Then in in figure 1, vertices $w_1,e_1$ can be dropped as a simplification step. Whereas in figures 2 and 3, the interior angle at $w_m\equiv e_1$ exceeds $2\pi$, which is forbidden. Remains figure 4: As the interior angle at $w_1\equiv e_k$ is already $2\pi$, we conclude that $w_1w_2$ is glued to $e_ke_{k-1}$. We can cut off triangle $ne_{k-1}e_k$ and paste it back in as triangle $nw_2w_1$, which is a simplification step that reduces the number of vertices.
Finally, assume that $nw_1$ is glued to $sw_m$ (and $se_1$ to $ne_k$). For figure 1, this is your second example and we are done. For figures 2, 3, 4, the interior angle at $w_1\equiv w_m$ is already $2\pi$, hence we must glue $w_1w_2$ to $w_mw_{m-1}$ (and in particular, $m\ge 3$). We can cut off $nw_1w_2$ and paste it back as $sw_mw_{m-1}$, which is a simplification step that reduces the number of vertices. (If $m=3$, this is the simplification procedure you gave as an example)
Next, assume that all vertices are equatorial and the perimeter is $e_1e_2\cdots e_me_1$ for some even $m\ge 2$. If $m=2$, vertices $e_1$ and $e_2$ must be antipodal and we can rotate the polygon to move the vertices to $n$ and $s$, bringing us to figure 1 above. If rotating equatorial vertices to the poles is frowned upon, we can of course also cut along the diagonal $e_1e_2$ passing through the pole in the interior of the polygon and reassemble the resulting spherical triangles accordingly.
So assume $m\ge 4$. In order to produce $S^2$, there must exist a vertex such that its two incident edges are glued together. Wlog $e_1$ is such a vertex, i.e., $e_1e_2$ is glued to $e_1e_m$. This brings the interior angle at $e_2\equiv e_m$ already to $2\pi$, so that we must glue $e_2e_3$ to $e_me_{m-1}$. We conclude that vertices $e_2$ and $e_m$ can be dropped, simplifying the polygon.
Finally, assume that only one pole is a vertex of the polygon, i.e., the perimeter is $ne_1\cdots e_mn$ for some odd $m\ge 3$, and we glue $ne_1$ to $ne_m$. In the resulting perimeter $e_1e_2\cdots e_m(\equiv e_1)$, there must be a vertex such that its incident edges are glued. If $e_1e_2$ is glued to $e_me_{m-1}$, we can cut off $ne_1e_2$ and reinsert it as $se_me_{m-1}$, which takes us back to the case of both poles used.
Hence we may assume that there is some $1<i<m$ such that $e_ie_{i-1}$ is glued to $e_ie_{i+1}$. This allows us to cut off $e_{i+1}\cdots e_mn$ (which may be degenerate) and reinsert it as $e_{i+1}'\cdots e_1n$. After that, cut off $ne_ie_{i+1}$ and reinsert it as $se_ie_{i-1}$, which takes us back to the case that both poles are used.