We know that there is a physical interpretation for symplectic manifolds (briefly, the fact that a sympletic form assigns to any Hamiltoninan a vector field which describes the motion of particles). My question is if there is a physical meaning to the fact that all symplectic forms locally look alike, i.e. they are standard locally,(for instance, for the cotangent bundle of a manifold, what does it mean that any nonstandard symplectic form locally looks like the standard one which basically imposes the Hamilton's equations on the hamiltonian H)?
Another way to interpret this question is: Is there a 'heuristic' reason that all closed nondegenerate $2$forms in $2n$ dimensions are locally equivalent? (It becomes more reasonable to ask this when you realize that nothing like this holds for, say, $3$forms in general dimension, even for $3$forms in dimension $5$.) There is a 'function count' heuristic reason for this local equivalence, which goes like this: If $\omega$ is a closed $2$form in dimension $n$, then, locally $\omega = \mathrm{d}\alpha$ for some $1$form, by Poincaré's Lemma. Now, in any local coordinates, $\alpha = a_i(x) \mathrm{d} x^i$, so $\alpha$ (and hence $\omega$) 'depends' on a choice of $n$ functions of $n$ variables, the $a_i$. However, local coordinate changes also depend on $n$ functions of $n$ variables, so you might expect that, by choosing coordinates cleverly, you could bring $\alpha$ into some standard form, just as you can put a (nonzero) vector field into a standard local form in socalled 'flowbox coordinates'. It turns out that you can't quite do this for $1$forms (another piece of evidence that you shouldn't mentally identify $1$forms with vector fields as people often do when they first start thinking about differential forms); there is an algebraic invariant, the rank $\rho(p)$ of $\mathrm{d}\alpha$ as a tensor at each point $p$. However, if this rank is locally constant, then, yes, you can choose coordinates so as to reduce $\mathrm{d}\alpha$ to a standard normal form. That is the content of Darboux' Theorem. To reduce $\alpha$ itself to standard form, you'd need a little more information. That is the content of the Pfaff Theorem. However, because $\alpha$ wasn't uniquely determined by $\omega$ (you could always add a term $\mathrm{d}f$ to $\alpha$), this extra information gets thrown away. (In fact, it is this 'gauge' ambiguity that shows that you should still have one arbitrary function left over in your choice of normalizing coordinates once you put $\omega$ in normal form. Hence the arbitrary function, i.e., the arbitrary choice of a Hamiltonian, in finding vector fields whose flows preserve $\omega$.) If you try to do the same count for, say, closed $3$forms in dimensions above $4$, you'll see immediately that there is no hope for such a simple normal form. 


Obviously, this is a question that could be interpreted in different ways. For me, Darboux's theorem is the symplectic analogue of the theorem that a flat Riemannian manifold (i.e. one where Riemann's curvature tensor vanishes) is locally the same as $\mathbb{R}^n$. For Darboux, the analogue of the Riemann curvature tensor is $d\omega$, the differential of the symplectic form. So one variation on your question is "why should the symplectic form on phase space be closed?" which has a very clear answer:
That is definitely not very physically realistic. The way to compute this is by taking the Lie derivative of $\omega$ by a Hamiltonian vector field $X_f$, using Cartan's magic formula $\mathcal{L}=d\iota+\iota d$. One finds that $$\mathcal{L}_{X_f}\omega=d(\omega(X_f,))+d\omega(X_f,,)=ddf+d\omega(X_f,,))=d\omega(X_f,,),$$ so closedness is essentially equivalent to invariance of $\omega$ under time translation. 


In Riemannian or semiRiemannian geometry, the reason that our spaces don't all look alike in a neighborhood of a point is that parallelism fails, and the extent to which parallelism fails can be described as a curvature, which is invariant. For example, if two geodesics both start from a point and later on intersect again at some other point, this is a sign of curvature. It's invariant because intersection is invariant. It doesn't matter what coordinates you pick  an intersection is an intersection. In general relativity, geodesics are the trajectories of test particles, and they can intersect in ways that imply curvature. In the phase space for a Hamiltonian system, the integral curves never intersect, which is physically because preparing a physical system in a definite state is supposed to determine its future time evolution according to the laws of physics. (As in general relativity, the curves can be the trajectories of particles through phase space, but a point in phase space doesn't just tell you the particle's position, it tells you its entire physical state.) Since the integral curves can't intersect, if someone draws a set of integral curves for you, you can always bend and stretch the picture so that the integral curves look like parallel lines  in some finite neighborhood. 

