I have to say first that I haven't read enough about vector bundles because currently I don't have time for it. But I hope that the question isn't that much trivial.
Why don't define the vector bundle of the $k$-manifold $M$ simply as follows?
It is the topological space $M\times \mathbb R^k$, endowed with the product topology and a projection $\pi: M\times \mathbb R^k \to M$, (which is continuous because it maps open sets to open sets,) and further we require that the fiber $\pi^{-1}(p)$, for $p \in M$, has the vector space structure of $\mathbb R^k$.
The local trivialisation is now simply $\pi$. Why isn't that enough ? what did I miss ?
(There are many different vector bundles on a given space, so I'm not sure if by "the vector bundle" you really mean the tangent bundle. In any case, the tangent bundle is an example of a vector bundle that doesn't fit your definition.)
Informally, the tangent bundle is the space of vectors "tangent to" or "inside" a manifold. This looks like a product space locally, but it can have a different global topological structure.
The hairy ball theorem demonstrates this difference. When $M=S^2$, there is no continuous map $M\to TM$ that assigns a nonzero vector to every point. If $TM$ were just a product space $M\times \Bbb R^k$, we could trivially assign a nonzero vector to each point using a map like $x\mapsto(x,(1,...))$. The tangent bundle captures the requirement that our tangent vector must vary continuously and remain tangent to $M$ as we move around on $M$.