Show that $\gamma'(0) = v.$ 
Let $M$ be a smooth manifold and $p \in M.$ Let $v \in T_p (M).$ Then there exixts a smooth curve $\gamma : (-\varepsilon,\varepsilon) \longrightarrow M$ with $\gamma (0) = p$ such that $\gamma'(0) = v.$

Let $\dim M = n$ and let $e_i$ be the $i$-th coordinate vector for $\mathbb R^n.$ Fix a chart $(\phi, U)$ in $M$ around $p.$ Then there exists an $\varepsilon \gt 0$ such that the $\varepsilon$-neighborhood of $\phi(p)$ is contained in $\phi (U).$ Now define $\gamma : (-\varepsilon, \varepsilon) \longrightarrow M$ by $$\gamma(t) = \phi^{-1} \left (\phi (p) + t \sum\limits_{i=1}^{n} v(x_i) e_i \right )$$ where $x_i$ is the $i$-th coordinate function corresponding to the chart $(\phi, U).$ Then it is claimed in my note that $\gamma'(0) = v.$ But I am having hard time showing this. What I need to show is that given $f \in C^{\infty} (p)$ (the space of all real valued functions defined on $M$ which are smooth at $p$) we should have $\gamma'(0) (f) = v(f).$ In other words for any $f \in C^{\infty} (p)$ we have $$\frac {d} {dt} \left (f \circ \gamma \right ) (t)\ \bigg \rvert_{t = 0} = v(f).$$
Any help would be appreciated.
 A: Recall that $\left \{\partial_1, \cdots, \partial_n \right \}$ is a basis for $T_p M$ and for any $v \in T_p M$ we have $v = \displaystyle\sum\limits v(x_i)\ \partial_i.$ Now let $\gamma'(0) = \displaystyle \sum c_i \partial_i.$ Then for all $i = 1, \cdots, n$ we have $$\begin{align*} c_i & = \gamma'(0) (x_i) \\ & = \frac {d} {dt} x_i \left (\phi^{-1} \left (\phi (p) + t \displaystyle v(x_i) e_i \right ) \right )\ \bigg \rvert_{t = 0} \\ & = \frac {d} {dt} \left (i\text {-th coordinate of}\ \left (\phi (p) + t \displaystyle \sum v(x_i) e_i \right ) \right )\ \bigg \rvert_{t = 0} \\ & = v(x_i) \end{align*}$$ This shows that $\gamma'(0) = v.$
A: We have :
$$v(f) = \sum_{i=1}^n v(x_i) \partial_i (f\circ \phi^{-1})|_{\phi(p)}$$
Also, $f\circ \gamma = (f\circ \phi^{-1}) \circ (\phi\circ\gamma)$, so by the chain rule :
$$\left.\frac{d}{dt}(f\circ\gamma)\right|_{t=0} = \left(\sum_i{\partial_i}(f\circ\phi^{-1})|_{\phi(p)} e_i^*\right)\cdot \left(\sum_j v(x_j)e_j\right) = \sum_i{\partial_i}(f\circ\phi^{-1})|_{\phi(p)}v(x_i) = v(f) $$
