line bundles over the circle I read in various places that up to isomorphism there are only two line bundles ( 1-d vector bundles) over a circle, the trivial one and the mobius strip. On the other hand, when I make a mobius strip from a strip of paper, I note that I can give the paper as many rotations as I want. No rotation of the paper and simply gluing the end pieces of the strip gives a cylinder without tops (which we can think as the trivial line bundle), one rotation gives the mobius strip. But more rotations of the paper are possible and they are not isomorphic to each other; you cannot make one change  to the other without tearing the paper.
Now I understand  that these are not line bundles as they are not extended to infinity and they are "bounded" together in a way that infinite fibers or vectors would not be. 
I am just wondering: Is there a notion of "bundles" here over a circle which extends to more than the trivial and mobius strip? After all, no rotation and one rotation of my strip do match up with the two possible vector bundles (trivial and mobius). Can we define a notion of "bundle" for more rotations of the strip?
EDIT:  The comments below basically mean that I have not understood the situation correctly. I will not delete the question since I think others might run into the same misunderstanding.  
 A: I will try to sum up the existing comments and talk about the linking number.
Isomorphism vs. Isotopy
So basically there are two things interplaying here when you talk about line bundles over $S^1$: isomorphisms and isotopies. Let's look at the definitions:


*

*Two line bundles $E, E' \xrightarrow{p, p'} S^1$ are said to be isomorphic if there exists a map $f : E \to E'$ that makes the triangle commute ($p' \circ f = p$), such that the induced map on fibers is linear (these two conditions make $f$ into a vector bundle morphism), and moreover $f$ has an inverse which is also a vector bundle morphism.

*Let two line bundles $E, E' \xrightarrow{p, p'} S^1$ and embeddings $E, E', S^1 \subset \mathbb{R}^3$. Then an isotopy between $E$ and $E'$ is a map $H : [0,1] \times \mathbb{R}^3 \to \mathbb{R}^3$ such that:


*

*$H(0, -)$ is the identity of $\mathbb{R}^3$;

*every $H(t, -)$ is a homeomorphism $\mathbb{R}^3 \to \mathbb{R}^3$;

*$H(t, -)$ is the identity when restricted on $S^1$;

*$H(1, E) = E'$ and $p' \circ H(1, -) = p$.



It's easy to see that if $E$ and $E'$ are isotopic in $\mathbb{R}^3$, then they are isomorphic. The converse is not true, as you noted: as band twisted twice is isomorphic to the trivial line bundle, but it is not isotopic in $\mathbb{R}^3$ to the cylinder. The isotopy class depends on the specific embedding in $\mathbb{R}^3$, but the isomorphism class does not.
Classification
Isomorphism classes: Stiefel-Whitned
In fact there are only two isomorphism classes, the trivial one and the one corresponding to the Möbius strip; see this question. Basically (this is a different version of the reasoning in the answers I linked), isomorphism classes of line bundles on $X$ are classified by the first Stiefel-Whitney class(*) $w_1(X) \in H^1(X; \mathbb{Z}/2\mathbb{Z})$, and $H^1(S^1; \mathbb{Z}/2\mathbb{Z}) \cong \mathbb{Z}/2\mathbb{Z}$.
Isotopy classes: Linking number
Now it is actually possible to classify (smooth?) isotopy classes too.
Let $E \to S^1$ be a line bundle embedded in $\mathbb{R}^3$, and let's assume that it is smooth for simplicity (the continuous case is more complicated). Then you can construct two curves out of it:


*

*The first one $\gamma_0$ corresponds to the zero section of the line bundle.

*For the second one, take a tubular neighborhood of the zero section in $E$ and take $\gamma_1$ to be one of the boundary components.
Finally consider the linking number: $$\operatorname{lk}(\gamma_0, \gamma_1),$$ which is well defined up to sign. This number basically corresponds to your intuitive notion of "number of twists": it's zero for the cylinder, one for the Möbius strip, two if you add one more twist... And it classifies smooth isotopy classes of smooth line bundles embedded in $\mathbb{R}^3$.

(*) The reason for that is one of my favorite facts in mathematics (the complex case one is just as amazing): $$\operatorname{Vect}^\mathbb{R}_1(X) = [X, \mathrm{Gr}_1(\mathbb{R}^\infty) ] = [X, \mathbb{RP}^\infty] = [X, K(\mathbb{Z}/2\mathbb{Z}, 1)] = H^1(X; \mathbb{Z}/2\mathbb{Z}).$$
A: Not a PROOF but a non-sufficient simple algebraic reason for twice twisted is same as trivial for further readers:
Define an equivalence relation on $\Bbb R^2$ by
declaring that $(x', y')\sim(x,y)\iff (x',y')=(x+n,(-1)^ny)$ for some
$n \in\Bbb Z$. This defines the Mobius bundle. To see that if we twist the band twice is same as trivial bundle (that is the identification $(x,y)\sim (x+m,y)$ for some $m$), just apply this identification once again, i.e.: (If I am not mistaken by notations)
$$\text{twisted }\Bbb R^2\ni(a', b')\sim(a,b)\in\text{twisted }\Bbb R^2\iff (a',b')=(a+n,(-1)^nb)=(x+2n,(-1)^{2n}y)=(x+2n,y)\iff (a', b')\sim(x,y).$$
that says $(z,w)\sim (z',w')\iff (z',w')=(z+2n,w)$ that is the trivial bundle. (first factor is a circle and second is a line)
$\qquad\qquad\qquad\qquad\qquad$Source: Wikipedia
