Why aren't the "higher twist" Möbius bands distinct line-bundles over $S^1$? It is well known (using for instance sheaf cohomology) that there exist only two possible one-$\mathbb R$-dimensional vector bundles over $S^1$: the trivial bundle and the Möbius bundle. But what about the Möbius bands with not one, but $n\in \mathbb N$ half-twists? It is not obvious to me why these are not distinct line bundles. I am also not entirely comfortable with sheaf cohomology arguments. 
This is also physically motivated; if you make a Möbius band, it's easy to see that two half-twists don't "cancel" to make a trivial bundle; so why do they "cancel" topologically? 
 A: As Roland explained in his comments, you have to distinguish between a line bundle $L$ over $S^1$ and embeddings of $L$ into $\mathbb R ^3$. In fact, any subspace $L' \subset \mathbb R ^3$ which is homeomorphic to $L$ is a geometric instantiation of $L$, and it may be impossible to deform (inside $\mathbb R ^3$) one instantiation $L'_1$ into another instantiation $L'_2$ without rupturing. The mathematical concept of an "admissible" deformation process inside an ambient space is isotopy.
Take for example the trivial line bundle $L = S^1 \times \mathbb  R$. Since $S^1 \subset \mathbb R ^2$, it is a genuine (unbounded) subset of $\mathbb R ^3$. However, there are many other instantiations of $L$, for example $L' = S^1 \times (-\frac{1}{4},\frac{1}{4})$ or twisted versions $L'_n$ of $L'$ (with $n$ full twists). All these are homeomorphic copies of $L$, but no two of them can be deformed into each other.
As an analogue you may also consider the circle $S^1$. A homeomorphic copy of $S^1$ in $\mathbb R ^3$ is a knot, and knot theory is a highly non-trivial field devoted to the classification of knots.
