$\DeclareMathOperator{diam}{diam}$
Suppose we have a convex spherical polygon $P$ and suppose that we've figured out the two vertices which are the greatest distance apart, say $u$ and $v$. Consider the double $DP$ of $P$, that is, the metric space defined by gluing two copies of $P$ along their boundary and measuring distances by shortest paths. Let's call the metric $d$. I want to calculate $\diam DP$. I'm happy to restrict to that case of triangles or quadrilaterals.
My guess is that, in fact, $\diam DP=d(u,v)$. This is because if we have two points $p$ and $q$ on the interiors of different copies of $P$, the shortest path joining them is made up of two segments ending on the same edge of $P$, and it seems to me that the sum of these two lengths could always be increased by moving one of the points away from that edge (I can see there are special cases where this doesn't work). So the diameter must be acheived within one copy of $P$, where it is clearly $d(u,v)$. But this argument doesn't seem very watertight.
In general, suppose we had a metric space $S$ built out of many copies of $P$ by gluing sides together in pairs by isometries. Let's also suppose that the sum of angles around any vertex is at most $2\pi$. How can we figure out $\diam S$?