An intuitive vision of fiber bundles In my mind it is clear the formal definition of a fiber bundle but I can not have a geometric image of it. Roughly speaking, given three topological spaces $X, B, F$ with a continuous surjection $\pi: X\rightarrow B$, we "attach" to every point $b$ of $B$ a closed set $\pi^{-1}(b)$ such that it is homeomorphic to $F$ and so $X$ results a disjoint union of closed sets and each of them is homeomorphic to $F$. We  also ask that this collection of closed subset of $X$ varies with continuity depending on $b\in B$, but I don't understand why this request is formalized using the conditions of local triviality.
 A: Maybe it is helpful to take a map $f\colon\mathbb R\to\mathbb R$ and consider its  graph $\Gamma(f):=\{(x,f(x)) |x\in \mathbb R\}$. We get a continuous map $\pi\colon \Gamma(f)\to \mathbb R$, $(x,f(x)) \mapsto x$. For every $p\in \mathbb R$, the preimage $\pi^{-1}(p)$ is again a single point, so this has a chance to be a fiber bundle. However, it is a fiber bundle, if and only if $f$ is continuous:
If $\pi$ is a fiber bundle near $x$, then there is a local trivialization, i.e. a homeomorphism $\pi^{-1}(U)\to U$ over $U$ for $U\subset \mathbb R$ a neighborhood of $x$. This implies that this homeomorphism is precisely  given by $\pi$ and so the assignment $x\mapsto (x,f(x))\mapsto f(x)$ (which is the inverse of the homeomorphism composed with the projection) is continuous near $x$ and hence $f$ is continuous near $x$. On the other hand, if $f$ is continuous, the map $\pi$ itself is already a homeomorphism (an inverse is given by $x\mapsto (x,f(x))$) and hence a fiber bundle.
So, local triviality in this case means that the map $f$ is locally continuous and the fiber which is essentially $f(x)$, varies continuously with the base point. Also, it is not hard to see that for the sign-function (as an example of a non continuous function), the non continuity at $0$ ruins the local triviality.
I hope this is a little helpful.
A: One example: A branched cover is a fiber bundle, where the fiber is a set of points.
http://www.math.cornell.edu/~hatcher/AT/ATch1.pdf
See the section "Representing Covering Spaces by Permutations", p. 68.  You can build a branched cover by covering your space with open sets $X \subset \bigcup U_i$  and your bundle with be covered by $U_i \times \{ 1, 2, \dots, n\}$  Then you need to consider what happens on $U_i \cap U_j $.  The transition function will be a bijective map 
$$\{ 1, 2, \dots, n \} \to \{ 1, 2, \dots , n \} $$
which is a permutation.
Therefore, branched covers can be thought of as fiber bundles over spaces where the fibers are finite sets.

By covering your space with open sets $X \subset \bigcup U_i$, taking direct products with your fiber $U_i \times V$ and considering what happens over interections $(U_i \cap U_j) \times V$ you can "patchwork" a fiber bundle together.

This construction is very general.  If your fiber $V$ is a vector space, your transitions maps are invertible linear maps
$V \to V$ which take values in the general linear group $GL(V)$.  So the set of vector bundles can be thought of as the space representations of the fundamental group $\pi_1(X) \to GL(V)$.
Then you can ask your transition functions be holomorphic and then it's a holomorphic vector bundle. Or you can ask your transition functions be continuous or have 5 derivatives or "reasonable" restriction (i.e. consistent with vector bundle axioms).

Alternatively, you can study a vector bundle by looking at its section, which are maps from the base space into the vector space.  In high school and college, we deal mostly with the trivial bundle $\mathbb{R}^2$ and where the base space is $\mathbb{R}$ and the fibers are $\mathbb{R}$.  Then we look at sections $$\{ (x, f(x)\} \in \mathbb{R}^2$$
In this way we can consider trivial bundles over the circle $S^1 \times \mathbb{R}$ and consider only those sections which are square-integrable i.e.
$$ \int_{S^1} |f(x)|^2 dx < \infty$$
Morally, this vector bundle is still the cylinder, but we are ruling out certain sections.
Question: What is the analogue of Fourier analysis for the Möbius band in this picture?

The sphere can be thought of as a fiber bundle over the line.  Indeed the fiber 
$$ \{ x^2 + y^2 + z^2 = 1 \} \cap \{ z = k\} = \{ (x,y,k): x^2 +y^2 = 1- k^2 \}$$
is a circle except at the endpoints $k = \pm 1$, where the fiber degenerate to points.  Also the torus is a fiber bundle $S^1 \times S^1$.
