I find a common rule that unites generation of Viviani Frill and the the Möbius Band.
$$ \phi =\theta $$
where $ \phi,\theta $ are spherical coordinates.
Please comment if this way looking at it appeals or otherwise. It has also to do with
bringing in higher dimensionality by conversion of a constant to a parameter/variable.


EDIT1: The adopted first one given here is same as Wiki. May be equivalent embedding by parameter/variable changes? The plot above shows Band inside Torus if 1 > u > -1. Three parametrizations:
$x(u,v)= \left(1+v \cos u\right)\cos u$; $y(u,v)= \left(1+v \cos u\right)\sin u$; $z(u,v)= v\sin u $;
Mathworld:
$x(u,v)= \left(1+v \cos \frac{u}{2}\right)\cos u$; $y(u,v)= \left(1+\,v \cos\frac{u}{2}\right)\sin u$; $z(u,v)= \,v\sin \frac{u}{2}$;
Wiki:
$x(u,v)= \left(1+\frac{v}{2} \cos \frac{u}{2}\right)\cos u$; $y(u,v)= \left(1+\frac{v}{2} \cos\frac{u}{2}\right)\sin u$; $z(u,v)= \frac{v}{2}\sin \frac{u}{2}$;
EDIT2:

The extended period also shows proper embedding for the parametrization:
$$ [(b + u \cos\theta ) \cos\theta, (b + u \cos\theta) \sin \theta, u \sin\theta] $$
Your picture of a Mobius band on a torus is incorrect...if you trace along the edges of the strip you've drawn, you'll see that it has two disjoint edges.
It CAN'T be a Mobius band, because if it were, then the strip perpendicular to it -- a neighborhood, in the torus, of the centerline --- would also have to be a Mobius strip, because their direct sum is a trivial bundle over the circle. But if there were a Mobius strip embedded in the torus, the torus would be nonorientable.
As for the V-frill and its relation to the M-band: I don't see much in anything that's this strongly coordinate-dependent, esp. when the coordinate system (on the sphere at least) is degenerate at the poles...but maybe I'm just missing something. This seems more like a statement about the cylinder, which admits an almost-everywhere nice map to the sphere (i.e., "contract the two end-circles to points") and an almost-everywhere-nice map to the torus (i.e., "identify the two end circles").