Why is $[\widetilde{v},\widetilde{w}]_p(f)=0$ when $f$ has a critical point at $p$? Let $M$ be a smooth manifold and $f$ a smooth function $M\to\mathbb{R}$.  Let $p$ be a critical point of $f$.  We define the Hessian of $f$ at $p$ to be the symmetric bilinear functional $f_{**}$ on $T_p M$ given by $f_{**}(v,w)=\widetilde{v}_p(\widetilde{w}(f))$ where $\widetilde{v}$ and $\widetilde{w}$ are extensions of $v$ and $w$ to vector fields on $M$.  It must be checked that $f_{**}$ is symmetric and well defined.  
Milnor states that $f_{**}$ is symmetric because $\widetilde{v}_p(\widetilde{w}(f)) - \widetilde{w}_p(\widetilde{v}(f)) = [\widetilde{v},\widetilde{w}]_p(f)$ where $[\widetilde{v},\widetilde{w}]$ denotes the Poisson bracket, which he says satisfies $[\widetilde{v},\widetilde{w}]_p(f)=0$ because $f$ has a critical point at $p$.  Thus since $\widetilde{v}_p(\widetilde{w}(f))$ is independent of the extension of $v$ and $\widetilde{w}_p(\widetilde{v}(f))$ is independent of the extension of $w$ we have that $f_{**}$ is well defined and symmetric by the preceding.
My confusion is why $[\widetilde{v},\widetilde{w}]_p(f)=0$.  I'm not familiar with Poisson bracket but this looks like the Lie bracket to me, not the Poisson bracket.  I'm also not sure why it is zero when evaluated at $p,f$.  
Also one more (non essential) question out of curiosity, can we only define the Hessian in a coordinate free way at points where a function has a critical point?  This would seem strange to me.  
 A: You're correct that $[\tilde{v}, \tilde{w}]$ is the Lie bracket of vector fields. $[\tilde{v},\tilde{w}]_p(f) = 0$ since
$$[\tilde{v},\tilde{w}]_p(f) = df_p([\tilde{v},\tilde{w}]) = 0,$$
as $p$ is a critical point of $f$.
This definition of the Hessian only makes sense at critical points. One way to get a coordinate-free definition of the Hessian everywhere on the manifold is to choose a Riemannian metric (so that we get the Levi-Civita connection $\nabla^{LC}$) and define
$$\mathrm{Hess}(f)(X,Y) = (\nabla^{LC} df)(X,Y).$$
This definition depends on the choice of Riemannian metric, however. Nevertheless, one can show that at a critical point, it is equivalent to the definition from Milnor's book.
A: Here I think is a complete answer to why [̃,̃]() = 0:  It is easy to check that the actions of ̃ and ̃  are derivations implies that the action of [̃,̃] is also a derivation (a linear functional which satisfies the product rule of elementary calculus).  Since all derivations are the action of some vector field (http://www.math.toronto.edu/mgualt/courses/17-1300/docs/17-1300-notes-12.pdf Theorem 4.15) the action of [̃,̃] is also that of a vector field.  Seth's remark of Jan. 11, 2016 now applies.
