Tangent space of Cotangent bundle at zero section? Let $M$ be a differentiable manifold with cotangent bundle $T^*M$.
How can I prove that $T_{(p,0)}T^*M$ is naturally isomorphic to $T_pM\oplus T_pM^*$?
If this true, then I think I could prove that the Hessian of $f\colon M\to \mathbb{R}$ is well-defined (I mean, without choice of Connection or Riemannian metric) at critical point of $f$. 
This is not any homework. 
 A: Things are much nicer i.e. more general and canonical than that!
a) Let $\pi:N\to M$ be a submersion of manifolds. From it you obtain the exact sequence of vector bundles on $N$: $$     0\to  T^{vert}(N) \to   T(N) \stackrel {d\pi}{\to} \pi^{-1}T(M)  \to   0  \quad (*)       $$ Profound, eh? Not at all!
This is just a fancy way of looking at the differential of the map   $\pi: N\to M.$
In order that both the tangent bundles to  $N$ and to  $M$ live on $N$, you have to pull back $T(M)$ to $N$: that is why we have $\pi^{-1} T(M)$ on the right.
The kernel is the set of vertical tangent vectors to $N$, those that lie along the level lines  $N[n]\stackrel {def}{=}\pi^{-1}(\pi (n))$ of $\pi$.
In other words at a point $n\in N$ the fibre  of $T^{vert}(N)$ is  $$T^{vert}_n(N)=T_n(N[n])   \quad (**)$$
b) Consider now a vector bundle $\pi:V\to M$ on $M$.
You then have the canonical exact sequence of vector bundles on $V\;$ (yes, vector bundles on a vector bundle!) :
$$   0\to  T^{vert}(V) \to   T(V) \stackrel {d\pi}{\to} \pi^{-1}T(M)  \to   0 \quad (***)$$
In this new set-up you have the interesting identification $ T^{vert}(V)=\pi^{-1}(V)$.
 This boils down to the fact that the tangent space at any point $e\in E$ of a vector space $E$ is that vector space itself: $T_e(E)=E.$
Hence $(***)$ becomes $$        0\to  \pi^{-1}(V) \to   T(V) \stackrel {d\pi}{\to} \pi^{-1}T(M)  \to   0 \quad (****)              $$
c) Finally, if you restrict this last exact sequence (****) to the zero section of $V$, identified to $M$, you get 
$$     0\to  V \to   T(V)|M \stackrel {d\pi}{\to} T(M)  \to   0 \quad   \quad (*****) $$
d) Although this is not very helpful, you can if the manifold $M$ is paracompact (see here) non-canonically split the exact sequence  (*****) and obtain the isomorphism $$ T(V)|M \cong  V \oplus T(M)$$ 
At a point $(m,0)\in M\subset V$ this gives the decomposition $$ T_{(m,0)}(V) \cong  V_m \oplus T_m(M)$$
and for $V=T^*(M)$ this is what you were looking for.
