Equivalence between function being Morse and $df$ being transversal to zero section. The proof I know of the fact that 

$f:M \to \mathbb{R}$ is Morse iff  $df:M \to T^*M$ is transversal to the zero section

uses local coordinates heavily. I would like to know if there is an intrinsic approach. Since both $df$ and $\mathrm{Hess}$ at a critical point admit fairly intrinsic definitions, I would hope so.
 A: I wasn't able to find an answer directly using the identifications I outlined in the comments (thanks to @TedShifrin for the comments!). Namely, I wasn't able to prove that $(D_vdf)_p:T_pM \to T^v_{(p,0)}T^*M \simeq T_pM^*$ is equal to $\widetilde{\mathrm{Hess}}_pf:T_pM \to T_pM^*$ without coordinates and directly. Therefore, I will leave the question open in hopes of someone doing that.
But with the advent of a connection on $M$, I was able to get more machinery in order to do this intrinsically more easily.
Recall that a connection on $M$ induces a connection $\nabla$ on tensors. We will now verify that at a critical point, $(\nabla df)_p=\mathrm{Hess}_pf$.
We can see that since
\begin{align*}
(\nabla df)_p(X,Y)=(\nabla_Ydf)(X)_p=Y_pXf-df_p(\nabla_{Y_p}X)=Y_pXf=\mathrm{Hess}_pf.
\end{align*} 
This computation also shows that the Hessian at a critical point indeed does not depend on the connection you take, if defined by $(\nabla df)_p$.
Note that this implies that $\widetilde{\mathrm{Hess}}_pf: \left(v \mapsto \widetilde{\mathrm{Hess}}_pf(v)\right)$ being an isomorphism is equivalent to $(\nabla df)_p: \left(v \mapsto (\nabla_vdf)_p\right)$ being an isomorphism.
Now, with the induced connection on the bundle $T^*M$, we get by definition of the covariant derivative that
$$(\nabla_v df)_p=\lim_{t \to 0}\frac{\tau_{(p,x(t))}^{-1}(df_{x(t) })}{t},$$
where $x(t)$ is a curve adapted to $v$ and $\tau$ is parallel transportation. This gives by definition a vector on $T_{(p,0)}^vT^*M$. It is not difficult to prove that this is precisely $(D_vdf)_p(v)$ (c.f. Paternain - Geodesic Flows, the part concerning the connection map - Definition $1.12$ up to Exercise $1.16$, but generalizing from $TM$ to a bundle $E$.). Thus, the Hessian being non-degenerate is equivalent to $(D_vdf)_p$ being an isomorphism. Since $(D_hdf)_p$ is the identity and when expanded to $T^*M$ is the derivative of the null-section, $(D_vdf)_p$ being an isomorphism is equivalent to $df_p$ being transversal to the null-section.
