$\newcommand{\R}{\mathbb{R}}$ $\newcommand{\M}{\mathcal{M}}$ $\newcommand{\Hs}{\operatorname{Hess}}$ $\newcommand{\al}{\alpha}$
This is a curious inquiry:
Let $f:\R^N \to \mathbb{R}^n$ be a smooth map, $N>n$, and let $c \in \R^n$ be a regular value of $f$. Then $\M=f^{-1}(c)$ is a submanifold of $\R^N$, and for $p \in \M$, $T_p\M=\ker df_p$.
Let $v \in T_p\M$ and let $w \in \R^N$ satisfy
$$ \operatorname{Hess}f_p\big(v,v\big)+df_p(w)=0. \tag{1} $$
Question: Does there exists a smooth (or $C^2$) path $\alpha:(-\epsilon,\epsilon) \to \M$, such that $\alpha(0)=p, \dot \alpha(0)=v, \ddot \alpha(0)=w$?
The converse direction holds. Indeed, differentiating $$ f(\al(t))=c, $$ one gets $$ df_{\al(t)}(\dot \al(t))=0. $$ Differentiating one more time, $$ \operatorname{Hess}f_{\al(t)}\big(\dot \al(t),\dot \al(t)\big)+df_{\al(t)}(\ddot \al(t))=0. \tag{2} $$ Plugging in $t=0$ one gets equation $(1)$.
Edit:
Here is a proof for the special case where $v=0$: We need to show that any $w \in \ker df_p=T_p\M$ can be realized as an acceleration of a path:
Well, given such $w$, let $\al:(-\epsilon,\epsilon) \to \M$ be such that $\alpha(0)=p, \dot \alpha(0)=w/2$. Define $\beta(t)=\al(t^2)$. Then $\dot \beta(0)=0, \ddot \beta(0)=w$.
A possible approach:
For a given $v \in T_p\M$, the space $$ X_v=\{w \in \R^N \, | \, \operatorname{Hess}f_p\big(v,v\big)+df_p(w)=0\} $$ is an affine space of dimension $\dim \ker df_p=\dim \M$.
Let $\mathcal{A}_v$ be the space of admissible accelerations of paths in $\M$ whose initial velocity is $v$. The analysis above shows that $\mathcal{A}_v \subseteq X_v$.
If we can prove in advance that $\mathcal{A}_v$ is an affine space whose dimension equals $\dim \M=\dim X_v$, then the containment $\mathcal{A}_v \subseteq X_v$ implies equality.
(This is similar to how one proves $T_p\M=\ker df_p$).
Edit 2:
Here is a proof that $\mathcal{A}_v \supseteq a_0 + \text{span}\{v\}$.
Given $\al(t)$, with $\dot \al(0)=v$, consider $\beta(t)=\al(\phi(t))$, with $\phi(0)=0, \phi'(0)=1$. Then $$ \dot \beta(0)=v,\,\,\, \ddot \beta(0)=\ddot \al(0) +\phi''(0)v. $$ Writing $a_0:=\ddot \al(0)$ we obtain $\dot \beta(0)=a_0+\phi''(0)v \in \mathcal{A}_v$. Since $\phi''(0)$ can be arbitrary, $\mathcal{A}_v \supseteq a_0 + \text{span}\{v\}$.
This approach exhausts the information that can be obtained from considering reparameterisations of given paths.