Funny I've had to do this occasionally and I always do the following: for each edge E, attempt to construct a clockwise loop from that edge by traversing each vertex and picking the edge that is "most clockwise" compared to the incoming edge. Converse for counter clockwise. This gives you, for each edge, the two loops the edge is a part of (if they exist).

It sounds horribly slow, but isn't because the loop finding process for one edge can mark all edges it traverses as either being on the same loop, or not having a loop, for the orientation its checking. These edges can be skipped henceforth.

This does not find all loops, only the smallest disjoint loops (as every edge is at most part of two disjoint loops). But basically results in a similar planar "face" partition the author describes.

Its not clear to me whether my approach does not work (perhaps the angle compares fail on a torus? I've never had to deal with torii), or simply isn't suited because it misses some loops, but I thought I'd share it anyway :-)