Show that $ T_{p}M $ is a vector space of dimension $ n $. Show that $ T_{p}M $ is an $ n $-dimensional vector space. 
Hint: Given
two tangent vectors $ v_1 $ and $ v_2 $ at $ p $ with corresponding curves $ \gamma_1 $ and $ \gamma_2 $, we can
“add” the corresponding curves in $\mathbb{R}^n$ and then move back to $M$:
$ \tilde{γ} = \varphi^{−1}\circ (\varphi\circ \gamma_1 + \varphi\circ\gamma_2 − \gamma (p)) $
defines a curve through $ p $ with tangent vector $ v_1+v_2 $. $ \dots\square $
Thank you.
 A: By definition, $T_pM$ is the set of all maps $\gamma :(-\epsilon, \epsilon) \to M$ such that $\gamma(0) = p$, with the equivalent relation that $\gamma_1 \sim \gamma_2$ if and only if 
$(\phi\circ \gamma_1)'(0) = (\phi\circ \gamma_2)'(0)$ 
for some coordinate chart $\phi$ containing $p$. This defines a map $\Phi : T_pM \to \mathbb R^n$ by $[\gamma] \mapsto (\phi\circ \gamma_1)'(0)$. The above equivalent condition shows that $\Phi$ is injective. You can also show that $\Phi$ is surjective by considering $\phi^{-1}(tw): w\in \mathbb R^n$. 
$\mathbb R^n$ obviously has a vector space structure. With the identification $\Phi$, what is the vector space structure on $T_pM$? 
Let $v_1, v_2\in \mathbb R^n $ be represented by $\gamma_1$ and $\gamma_2$ (via $\Phi$) respectively. Define as in your question the curve
$\gamma = \phi^{-1}(\phi\circ \gamma_1 + \phi\circ \gamma_2)$.
Then $\gamma$ is a curve such that $\gamma(0) = p$ (We assume $\phi (p) = 0$) and 
$(\phi\circ \gamma)'(0) = (\phi\circ \gamma_1)'(0) + (\phi\circ \gamma_2)'(0)= v_1 + v_2$.
Thus $\gamma$ corresponds to the vector $v_1+v_2$ (via $\Phi$).
