A couple of related ideas:
The parity of the number of zeros of a transversal section of the Mobius strip bundle
Any smooth section of the Mobius strip, viewed as a line bundle over the circle, that has finitely many zeros and is transversal* to the zero section must have an odd number of zeros. If the number were even the bundle would be trivial. (proof?) and the total space of the bundle would be a cylinder not a Mobius strip. The parity of the number of zeros tells you whether the bundle has a non-zero section.
* Two submanifolds of a manifold,M, intersect transversally if at each intersection point their tangent spaces span the entire tangent space to M. In the case of the Mobius band the two submanifolds are the base circle and the section. Their tangent spaces at intersection points intersect only at the origin and span the two dimensional tangent space of the Mobius band at the point of intersection.
Circles on the Mobius strip viewed as sphere bundles over the circle
The circles that are parallel to the equator of the Mobius strip are twice as long as the equator and cover the equator twice when projected onto it. Two points on these circles lie above each point on the equator.
From the point of view of line bundles this is no surprise since in each fiber one can choose two points that are the same distance to the equator. In general, on any vector bundle with a Riemannian metric, points in each fiber that are a fixed distance to the origin of the fiber form a sphere and this creates a new fiber bundle, not a vector bundle, called a sphere bundle. A parallel circle of the Mobius strip is a 0 dimensional sphere bundle over the base circle.
An equivalent way to see that the Mobius strip bundle is non trivial - equivalent to showing that every section must have a zero - is to notice that the sphere bundle is connected. It is a single circle. If the bundle were trivial the sphere bundle would be split into two circles.
Finally it should be no surprise that the Mobius strip with its equator cut out is also connected (since each parallel circle is connected) and is not separated into two pieces as is a cylindrical strip.
Prove that the Mobius band minus its equator is the Cartesian product of a circle with a line segment. That is: it is homeomorphic to a cylinder. This means that a second cut along a circle will separate the strip into two pieces.
The Tangent Bundle to the Mobius strip
The tangent bundle to the Mobius strips splits into the direct sum of two lines bundles, one the lines tangent to the equator and the parallel circles to it, call it E, and the other tangent to the fiber lines of the Mobius line bundle, call it F.
A few things to think about:
The line bundle, E, is trivial.
The line. bundle ,F, restricted to the equator is just the Mobius bundle back again.
The line bundle,F, away from the equator is trivial.
If one identifies the boundary circle of the Mobius strip to itself to make a Klein bottle, show that these two line bundles extend to the Klein bottle. So the tangent bundle to the Klein bottle is the direct sum of two line bundles one of which is non-trivial. How does the non-trivial direct summand imply that the Klein bottle is not orientable?