It's pretty easy to do it in 3D. Doing anything on a 2D projection is hard, because the space is so distorted. When you can easily measure distances, this is a lot easier.

It's a little weird that their results are inside the sphere. I don't fully understand the equation, but it looks to me like it's using angles/rotations from a to the centre, so I would expect it to remain on the surface. Maybe the problem is just numerical error?

Of course, if they're projecting the solution back to the surface anyways, the simplest way would be to find the centre of the triangle in xyz space, ignoring the sphere, then projecting the result back to the surface. As long as the triangle isn't larger than half the circumference, the shortest distance through the sphere would correspond to the shortest distance on the surface.

To be honest, I'm more surprised they projected to a plane in the first place. If he was doing it all from scratch, it would probably be simpler to do the triangulation in 3D.