Tu's Introduction to Manifolds - Definition of vector bundle I'm trying to figure the motivation behind the necessaries pieces for the definition of vector bundles.
From section 12.3 of Tu's Introduction to Manifolds we have these definitions

A surjective map $\pi : E \to M$ of manifolds is said to be locally trivial of rank $r$ if : (i) each fiber $\pi^{-1}(p)$ has the structure of a vector space of dimension $r$. (ii) For each $p \in M$ there're open neighborhood $U$ of $p$ and a fiber-preserving diffeomorphism $\phi : \pi^{-1}(U) \to U \times \mathbb{R}^r$ such that for every $q \in U$ the restriction
  $$
\left. \phi \right|_{\pi^{-1}(q)} : \pi^{-1}(q) \to \left\{ q \right\} \times \mathbb{R}^r
$$
  is a vector space isomorphism. Such an open set $U$ is called a trivializing open set for $E$ and $\phi$ is called a trivialization of $E$ over $U$. 

Definition (i) is clear, definition (ii) a bit less, because I don't really understand where this triviliazation comes from, but I assume the term should hint something actually.
Two more definitions that I believe clarification might be useful as well

Given any map $\pi : E \to M$ we call the inverse image $\pi^{-1}(p) := \pi^{-1}(\left\{ p \right\})$ of a point $p \in M$ the fiber at $p$.

Isn't the fiber essentially just the preimage of this map $\pi$?

For any two maps $\pi : E \to M$ and $\pi' : E' \to M$ with the same target space $M$, a map $\phi : E \to E'$ is said to be fiber-preserving if $\phi(E_p) \subset E_p'$ for all $p \in M$.

why does the fiber preserving map requires $\phi(E) \subset E'_p$ and not $\phi(E) = E'_p$ which sounds a bit more natural to me?
 A: Consider two vector bundles on the unit circle: With $S^1 \times \Bbb R$ each fiber can be visualized as being perpendicular to the plane of the circle, and so the manifold easily embeds in $\Bbb R^3$ as an infinite cylinder. The product topology is exactly the same topology as the embedded cylinder.
But you know there is that other way of joining the ends of strip of paper: by giving a half-twist before joining the ends, you get a Möbius strip. The Möbius is topologically distinct from the cylinder: remove any simple loop from the cylinder, and you divide it into two parts. But there are loops on the Möbius strip which do not divide it.
So instead of topologizing $S^1 \times \Bbb R$ with the product topology and getting a cylinder, we can give it the topology of the Möbius strip. This is a different topological space, and therefore, a different manifold. It is not orientable, while the cylinder is.
Given some manifold $M$, the easy to create a vector bundle of fiber-dimension $k$ over it is to just use $M\times V$ for some $k$-dimensional topological vector space $V$ - in particular, $\Bbb R^k$. Because this is easy, and its topology and other relationships are obvious, this is called a "trivial bundle". But as with the Möbius strip, bundles need not be trivial.
However, we want it to be simple locally. Thus part (ii) of the definition means that vector bundles must be locally trivial. There is some neighborhood of each point in the base such that the restriction of the bundle to that neighborhood is trivial.
