I'm having some trouble solving this exercise from O'Neill's Semi-Riemannian Geometry with Applications to Relativity and I would like to find a satisfactory solution to it.
Let $M$ be an $n$-dimensional smooth manifold and $f : M \longrightarrow \mathbb{R}$ a differentiable function. Let $p \in M$ be a critical point of $f$, that is a point in $M$ such that $df_{p} =0$. Then, it is easy to see that there exists a Hessian function $H : T_{p}M \times T_{p}M \longrightarrow \mathbb{R}$, which is a symmetric bilinear form, given by $H(v,w) = \tilde{v}_{p}(\tilde{w}f)$, where $\tilde{v}$ and $\tilde{w}$ are smooth vector fields on $M$ such that $\tilde{v}_{p}=v$ and $\tilde{w}_{p} =w$ for any $v,w \in T_{p}M$. The exercise asks to show that if $v \in T_{p}M$ is such that $\alpha '(0) = v$ (where $\alpha : I \longrightarrow M$ is a differentiable curve from an open interval $I$ of $\mathbb{R}$ into $M$ such that $\alpha (0) =p$), then $H(v,v) = \left. \frac{d^{2}(f \circ \alpha )}{ds^{2}} \right\vert_{s=0} = (f \circ \alpha)''(0)$.
I'll describe my attempts to solve this problem and the trouble I've found while solving it just in case they lead to the desired solution.
The definition of $H$ does not depend on the smooth vector field that extends $v$ so I think this should be solved by taking an appropriate smooth field extending $v$. I believe that I was able to prove that if $\Phi : U \longrightarrow \Phi(U)$ is a coordinate chart on $p$ (where $U$ is an open subset of $M$ containing $p$ and $\Phi(U)$ is an open subset of $\mathbb{R}^{n}$) such that $\Phi(p)=0$, then there exists a bump function $\phi : M \longrightarrow \mathbb{R}$ satisfying that $\text{supp} \phi \subseteq U$ and $\phi \equiv 1$ on $V$ where $V$ is an open neighborhood of $p$ in $M$ which is contained in $U$. Then, if $x_{1}, \ldots , x_{n}$ are the coordinates of $M$ with respect to $\Phi$, we know that $v = \sum_{i=1}^{n} v(x_{i}) \left. \frac{\partial}{\partial x_{i}} \right\vert_{p}$, so I thought about defining the smooth vector field extending $v$ as $\tilde{v}_{q} = \sum_{i=1}^{n} \phi(q) v(x_{i}) \left. \frac{\partial}{\partial x_{i}} \right\vert_{q}$ for any $q \in U$ and $\tilde{v}_{r} = 0$ for any $r \in M \setminus U$. Since $v = \alpha '(0)$ and $\phi \equiv 1$ on an open neighborhood of $p$ contained in $U$, a direct calculation shows that: $$ H(v,v) = \left. \frac{d(\tilde{v}f \circ \alpha )}{dt} \right\vert_{t=0}$$ $$\qquad = \lim_{t \rightarrow 0} \frac{\tilde{v}_{\alpha (t)}f - (f \circ \alpha)'(0)}{t}$$ $$\qquad\qquad\qquad= \lim_{t \rightarrow 0} \frac{\sum_{i=1}^{n} \left. \frac{d(f \circ \alpha )}{ds} \right\vert_{s=0} \left. \frac{\partial}{\partial x_{i}} \right\vert_{\alpha (t)} - (f \circ \alpha)'(0)}{t}$$
The sum $\sum_{i=1}^{n} \left. \frac{d(f \circ \alpha )}{ds} \right\vert_{s=0}$ would be equal to $(f \circ \alpha )'(t)$ if the derivatives $\left. \frac{d(f \circ \alpha )}{ds} \right\vert_{s=0}$ were evaluated at $t$ instead of $0$ and this problem has all to do with the way I defined the smooth vector field $\tilde{v}$. This issue would be solved if $\alpha$ was an integral curve of $\tilde{v}$, but there's no guarantee that's the case. We know that $\tilde{v}$ has a unique integral curve $\beta$ such that $\beta(0) =p$, but $\alpha$ is an arbitrary curve whose derivative at zero is $v$.
So, is the curve $\alpha$ the integral curve of a smooth vector field extending $v$?
Can I define a smooth vector field extending $v$ in such a way that (for $t$ in a sufficiently small neighborhood of $0$) $\tilde{v}_{\alpha(t)}f = (f \circ \alpha)'(t)$?
Can I actually take a curve, say $\gamma$, which is an integral curve of a smooth vector field extending $v$ and prove in some way that $(f \circ \alpha)''(0) = (f \circ \gamma)''(0)$ (this could make some sense since $v= \alpha '(0) = \gamma '(0)$ and $v$ can be defined as the equivalence class of differentiable curves whose derivative at $0$ is the same)?
Are my calculations wrong and is it actually true that for the vector field $\tilde{v}$ defined above $H(v,v)=(f \circ \alpha)''(0)$?
My main question is finding a satisfactory solution to this exercise and a positive answer to the above questions would pretty much solve it. Or perhaps there is another approach to this exercise which would also be more than welcome.
Sorry for the long question, but any insight, hints, or ideas would be more than welcome and greatly appreciated. Thanks in advance.
EDIT 1: After doing some calculations, I think a better approach would be to take everything to functions in $\mathbb{R}$ and $\mathbb{R}^{n}$. Let $\varphi : W \longrightarrow \varphi(W)$, $W \subseteq \mathbb{R}^{n}$, be a coordinate chart on $p$. If $x_{1}, \ldots ,x_{n}$ are the coordinates of $M$ with respect to $\varphi$ and $u_{1}, \ldots , u_{n}$ are the usual coordinates of $\mathbb{R}^{n}$, then
\begin{equation} (f \circ \alpha )''(0) = (f \circ \varphi \circ \varphi^{-1} \circ \alpha)''(0) = \end{equation} \begin{equation} \sum_{i=1}^{n} \left( \sum_{j=1}^{n} \left. \frac{\partial^{2} (f \circ \varphi)}{\partial u_{j}u_{i}} \right\vert_{0} (u_{j} \circ \varphi^{-1} \circ \alpha)'(0) (u_{i} \circ \varphi^{-1} \circ \alpha)'(0) + \left. \frac{\partial (f \circ \varphi)}{\partial u_{i}} \right\vert_{0} (u_{i} \circ \varphi^{-1} \circ \alpha)''(0) \right). \end{equation} Using the bilinearity of $H$, we can find that:
\begin{equation} H(v,v) = \sum_{i=1}^{n}\sum_{j=1}^{n} (x_{i} \circ \alpha)'(0) (x_{j} \circ \alpha)'(0) \frac{\partial^{2}f}{\partial x_{i}x_{j}}. \end{equation} So we would be done if $\sum_{i=1}^{n} \left. \frac{\partial (f \circ \varphi)}{\partial u_{i}} \right\vert_{0} (u_{i} \circ \varphi^{-1} \circ \alpha)''(0) =0$ and from the fact that $p$ is a critical point of $f$, we know that:
\begin{equation} df_{p}(\alpha'(0)) = (f \circ \alpha)'(0) = (f \circ \varphi \circ \varphi^{-1} \circ \alpha)'(0) = \end{equation} \begin{equation} \sum_{i=1}^{n} \left. \frac{\partial (f \circ \varphi)}{\partial u_{i}} \right\vert_{0} (u_{i} \circ \varphi^{-1} \circ \alpha)'(0) =0, \end{equation}
which is almost what we want, but not quite. Any ideas on how to justify the equality $\sum_{i=1}^{n} \left. \frac{\partial (f \circ \varphi)}{\partial u_{i}} \right\vert_{0} (u_{i} \circ \varphi^{-1} \circ \alpha)''(0) =0$.
EDIT 2: I think I've got it. Following the notation of the first edit, since $p$ is a critical point of $f$, then $0$ is a critical point of $f \circ \varphi$. Consequently:
\begin{equation} \sum_{i=1}^{n} \left. \frac{\partial (f \circ \varphi)}{\partial u_{i}} \right\vert_{0} (u_{i} \circ \varphi^{-1} \circ \alpha)''(0) = d(f \circ \varphi)_{0}((\varphi^{-1} \circ \alpha)''(0)) = 0 \end{equation}
which finishes the proof. Could somebody confirm if this is actually right so I could post Edits 1 and 2 as an answer to the question?