What's the intuition behind the tangent bundle?

Well, when we work with a smooth manifold $M$ we can associate with each point $p\in M$ a vector space $T_p M$ of all vectors at $p$ tangent to $M$: this is the space of linear functionals obeying liebniz rule in the algebra of germs of functions at $p$.

This definition of vector is very intuitive, since it generalizes the main property of vectors in $\mathbb{R}^n$ of producing the directional derivative. Now, then we usually say: "well, we must find a way to assemble all of the tangent spaces together to have a domain and range for the derivative", then we define $TM$ as the disjoint union of all $T_p M$ and if $\pi : TM \to M$ is the projection on the first coordinate, we want to construct a vector bundle $\pi : TM \to M$.

Now, what doesn't seem clear to me is why do we want the structure of a vector bundle. The definition of a fiber bundle is meant as I understand, to make a space that locally looks like a product space, but why do we want this? Is this because $T\mathbb{R}^n = \mathbb{R}^n \times \mathbb{R}^n$ and we want to ``copy'' this behavior locally?

Also, how do we know that there's an obstruction in general to write $TM = M \times \mathbb{R}^n$? I've seem a question like this before here and there were answers based on hairy ball theorem and so on. The point is, this result needs that we first define $TM$ as it is define. If we don't know any of these theorems, how do we know that writing $TM$ that way is not possible?

Thanks very much in advance for the aid!

The bundle structure on the set $TM$ explains how it is a manifold in its own right, since one can construct by the bundle's local trivialization bundle charts, which map open sets of $TU\subset TM\to U\times\mathbb{R}^m$. It is especially useful for calculating objects on the (tangent) bundle, e.g. the Levi-Civita connection, in local coordinates where one needs local trivializations.