What does "locally trivial" do for us?

For the first question, if the dimension of $E$ is greater than or equal to the dimension of $M$, then having maximal rank is the same as the differential map being surjective. The other direction is also obvious.

Since we are dealing with differentiable manifolds and we have that property on the differential map, the manifold $E$ has a local product structure, i.e. for any $e\in E$ there is a neighborhood $U$ of $e$ and a manifold $F_e$ such that $U$ is diffeomorphic to $\pi(U)\times F_e$ (this is a direct consequence of the implicit function theorem). However, this product structure is not restrictive enough and you might end up, for example, with fibres with different topologies or homotopy types, for example the fibred manifold $(\mathbb{R}^2-\{0\},pr,\mathbb{R})$ where $pr:\mathbb{R}^2-\{0\}\rightarrow\mathbb{R}$ is the projection to the first coordinate, illustrates that phenomenon (by the way, this is also an example of a non-proper map which is not a fiber bundle). So, with only that product structure is hard to relate the topology of the manifolds involved. However, once you impose the local trivialization condition, you get rid of those "pathologies" (for instance the fibres on each component of the base become diffeomorphic). Also, local trivializations give the fibre bundle the homotopy lifting property, so you can compute invariants of the manifolds via exact sequences of homotopy groups, or the Serre spectral sequence, etc.

I'm sure that local trivializations do much more for us, but that's what I know so far.

Another fibred manifold which is not a fiber bundle is the following (non-proper) map: define $\mathbb{R}^2/\mathbb{Z}^2\rightarrow\mathbb{R}$ by $(x,y)$ mod $\mathbb{Z}^2\mapsto y-\sqrt{2}x$. Then this is a surjective sumbersion onto its image which is not locally trivial.