Let $f:M\to\mathbb{R}$ be a smooth function and $p\in M$ is a critical point of it.
The Hessian of $f$ at a critical point $p$ is a symmetric bilinear form $\operatorname{Hess} f_p$ s.t. $\forall v,w\in T_pM$, $$\operatorname{Hess} f_p(v,w)=V_p(W(f)),$$ where $V,W$ are the extensions of $v$ and $w$ to vector fields such that $V_p=v$ and $W_p=w$. Let the critical set of $f$ contains a submanifold $C$. Put a Riemannian metric on $M$ and $\forall p\in C$ consider the decomposition $$T_pM=T_pC\oplus T^{\perp}_pC.$$ Let $v\in T_pC$ and $w\in T^\perp_pC$. Then show that $$\operatorname{Hess} f_p(v,w)=0.$$
I would appreciate any comment
Let $\gamma:(-\epsilon,\epsilon)\to C$ represent the tangent vector $v$. That is, $\gamma(0)=p$, and $\dot{\gamma}(0)=v$. Let $W$ be any extension of the tangent vector $w$ to a neighborhood of $p$. Then we have $$W_{\gamma(t)}(f)=0$$ for every $t\in(-\epsilon,\epsilon),$ as $\gamma(t)$ is a critical point. By your definition of the Hessian, it now follows immediately that $\mathrm{Hess}f_p(v,w)=0.$
Note that the assumption $w\in T_p^\perp C$ is irrelevant here. This makes sense, as the Hessian of a function at a critical point is well-defined without any Riemannian metric or a connection (as implied by your definition).