When can we specify an everywhere orthogonal grid on a surface? Is there a classification for surfaces on which we can specify a grid that forms right-angles at every intersection without singularities? For example, we can do this on the plane, cylinder, torus, and mobius-strip, but not the cone nor sphere which (I think) will always have singularities when we try. It seems related to the idea of a developable surface but not quite.

Do such surfaces have any special importance for this fact? I understand that coordinates are, by nature, quite arbitrary, but this is a question about the existence of coordinates with a certain property, so perhaps it can be more fundamentally related to the topology of the surface. Is there a formalization of this concept in terms of a manifold's atlas? Or perhaps something about an everywhere diagonal metric-tensor? (Just spitballing).
Edit: The reason the cone fails is obvious: it isn't a regular surface / manifold, i.e. it has a "sharp point." Meanwhile, the reason the sphere fails might be a result of the Hairy-Ball Theorem? I'm beginning to think that finding an orthogonal grid doesn't make a manifold special. What would make a manifold special is proving that no such grid can exist, which may require a Hairy-Ball-like theorem. Still open to answers!
 A: Here's an answer for finite type oriented surfaces by which I mean any surface $S$ that is diffeomorphic to an oriented surface $F$ which is closed --- meaning compact, connected, and empty boundary --- minus a finite subset $A \subset F$. (I suspect something similar works in the nonorientable case, but I think I"ll restrict my answer to the orientable case)
First, if $A = \emptyset$ and hence $S=F$ is closed then a necessary and sufficient condition is that the genus equals $1$.
I claim that if $A \ne \emptyset$ then $S$ has such a grid as you ask for.
First let me describe all genus $0$ examples, i.e. all examples where $F$ is a sphere minus $1$ or more points. If $A$ has just $1$ point then $S=F-A$ is diffeomorphic to $\mathbb R^2$, in which the horizontal/vertical lines give such a grid, as depicted in your question. If $A$ has $k \ge 2$ points then $S=F-A$ is homeomorphic to $\mathbb R^2$ minus $k-1$ points, and again the horizontal/vertical lines (punctured at $k-1$ points) gives such a grid.
Also one can do all genus $1$ examples, i.e. where $F$ is a torus, because $F$ itself has such a grid, and then you can remove any finite subset of points.
You can see from these examples that if the genus is fixed and if you can describe a grid on the once punctured surface of genus $g$, then you obtain a grid on a genus $g$ surface punctured at any nonempty finite set.
So, what does one do if $F$ is the closed surface of genus $2$, and $S$ is $F$ minus $1$ point? In this case, consider a regular octagon in the plane, and glue every side with the opposite parallel side, using a gluing map identifies each opposite pair of sides by a Euclidean translation. The quotient under this gluing is a closed surface of genus $2$. All 8 vertices of the octagon are identified to a single point of $F$ which we take to be $A$. Now puncture $F$ at that one point to give the surface $S=F-A$, so we can think of $S$ as the quotient of the octagon with its 8 vertices removed. What one observes is that if one restricts the horizontal/vertical lines of $\mathbb R^2$ to the octagon with vertices removed, and then maps those lines to the quotient surface $S=F-A$, the result is a grid as you desire.
And then this generalizes to any genus: take a regular planar $4g$-gon, remove its vertices, and glue opposite side pairs; the horizontal\vertical grid on the plane descends to a grid such as you ask for, on the genus $g$ surface puncture at $1$ point.
