Let $a: \mathbb{R}^2 \rightarrow M$ a differential map such that $a(s, 0) = p$ for all $s \in \mathbb{R}$. Let $\gamma : \mathbb{R} \rightarrow T_p M$ the path given by $\gamma(s) : = \frac{\partial}{\partial t}_{|t=0} a(s,t)$. We identify $T_{\gamma(0)}(T_pM) $ canonical with $T_pM$. The path $\gamma$ define a tangent vector $\gamma '(0) \in T_p M$. Let $\phi \in \mathcal{E}(p)$, where $\mathcal{E}(p)$ is the set of the function germ in $p$ . Show that \begin{align} \gamma '(0)(\phi) = \frac{\partial ^2}{\partial s \partial t}_{|s=t=0} \phi(a(s, t)) \end{align} Hint: wlog $(M, p ) = (\mathbb{R}^n , 0)$ and $\phi$ linear
I have no idea how to approach this exercise. Any suggestions? Thanks in advance!
You said you have trouble understanding the notation $\gamma'(0)(\phi)$. That's a good place to start. Remember that elements of $T_pM$ act on $\mathcal{E}(p)$ like this: if $\gamma$ is a path in $M$ with $\gamma(0) = p$, then $\gamma'(0) \in T_pM$. If $\phi \in \mathcal{E}(p)$, the action is $$ \gamma'(0)(\phi) = \left.\frac{d}{ds}(\phi(\gamma(s)))\right|_{s=0} $$ In your problem, $\gamma(s) = \left.\frac{\partial}{\partial t} a(s,t)\right|_{t=0}$.