I'm talking of ordinary tori in 3D (donuts). According to a paper I read and my understanding, "the" square torus is the torus with $R/r = \sqrt{2}$ (where $R$ and $r$ are the major and the minor radii).
Why is it called "square"?
Is is true that two Villarceau circles on the square torus meet at a right angle?
If this is true, do you have a reference for this fact? I googled with no luck.
Speaking holomorphically, a torus is obtained from quotienting the Euclidean plane by a lattice. Two tori are equivalent if and only if their lattices are similar (one is a non-zero complex multiple of the other).
A square torus comes from the lattice of Gaussian integers, $\mathbf{Z} + i\mathbf{Z}$, or another square lattice.
Assume $a$ and $b$ are positive real numbers satisfying $a^{2} + b^{2} = 1$. A rectangular torus (i.e., whose lattice is rectangular) with aspect ratio $b/a$ may be obtained in two steps: Parametrize the flat torus in the three-sphere $$ T(u, v) = (a\cos u, a\sin u, b\cos v, b\sin v), $$ then map to Euclidean three-space by stereographic projection from $(0, 0, 0, 1)$: $$ X(u, v) = \frac{1}{1 - b\sin v}(a\cos u, a\sin u, b\cos v). $$ The flat torus parametrization is a local isometry from the Euclidean plane and stereographic projection from the three-sphere is conformal, so this parametrization is conformal.
If $a = b = 1/\sqrt{2}$, the preceding formula is a conformal parametrization of a square torus.
Villarceau circles are images of lines of slope $\pm1$ in the Euclidean plane, so yes, they meet at right angles.
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You may enjoy this interactive web program for drawing on a square torus. (Usage instructions are at the bottom of the linked page.) Particularly, you can test the claim about Villarceau circles by drawing diagonal lines.