Proving that the tangent space, considered as a set of equivalence classes of contours, is a vector space. [Note: Before you simply paste a link, I've already read several other similarly-phrased posts on this site, none of which quite answer my question or phrase it in a way I understand.]
Wikipedia defines the tangent space $T_xM$ of an $n$-manifold $M$ at $x\in M$ as follows.  Let $\varphi:U\rightarrow \mathbb{R}^n$ a chart, where $U\ni x$ is an open subset of $M$, and define
$$X\varphi_x(\gamma):=\frac{d}{dt}\left[\varphi\circ \gamma\right]\Big\rvert_{t=0}$$
over the set $C_{U,x}$ of contours $\gamma:I_{\gamma}\rightarrow U$, $I_{\gamma}$ a sufficiently small interval around 0, such that $\gamma(0) = x$ and $\varphi\circ\gamma$ is smooth.  Also, wlog, assume that $\varphi(x) = 0$ for simplicity.
Then $T_xM$ is the set $C_{U,x}$ modulo $\sim$, where
$$\gamma_1\sim\gamma_2\quad\Leftrightarrow\quad X\varphi_x(\gamma_1) = X\varphi_x(\gamma_2).$$
(It is straightforward to show that this definition does not depend on the choice of $\varphi$ or $U$.)
Now, Wikipedia says that $X\varphi_x(\gamma)$ induces a bijective map from $T_xM$ to $\mathbb{R}^n$, which gives a vector space structure to $T_xM$.  I understand why the map is well-defined and injective, but I do not understand how to prove surjectivity without using the fact that $T_xM$ is a vector space, a fact that (so Wikipedia suggests) should follow from surjectivity of $X\varphi_x$ and not the other way round.  Here's my approach so far:
Define $\gamma_k(t):= \varphi^{-1}(t\cdot e_k)$, where $e_k$ is the unit vector in $\mathbb{R}^n$ with a $1$ as its $k^{th}$ component.  Then $$X\varphi_x(\gamma_k[t]) = e_k.$$
If $T_xM$ is defined as the free vector space with base $\{\gamma_1, \dots, \gamma_n\}$ and $X\varphi_x:T_xM\rightarrow\mathbb{R}^n$ is assumed to be a linear map, then the conclusion that $X\varphi_x$ is surjective follows from the universal property of free modules.  But

*

*as I write above, this seems like putting the cart before the horse, proving that the map is surjective from the fact that $T_xM$ is a vector space and not v.v.; and

*even if this approach is correct, it's not clear that we are allowed to assume that $X\varphi_x$ is linear as a function of $T_xM$ (unless we extend it to be the unique homomorphism guaranteed by the universal property).

*indeed, why does the map $X\varphi_x$ matter at all if we can show that $T_xM$ is a vector space another way?

What am I missing?  Can I prove that $T_xM$ is a vector space using only the tangent vector definition (i.e., without also using the definition in terms of derivations)?
 A: Let $p \in M$ be a point in $M$ and let $(U,\varphi)$ be a chart around $p$. One can endow $T_p M$ with a vector space structure as follows:

*

*Scalar multiplication:
Let $[\gamma] \in T_p M $ and $\lambda \in \mathbb{R}$, then define a new curve  $\overline{\gamma }: (-\delta,\delta)  \rightarrow M; t \mapsto \gamma(\lambda t)$. Since $(\varphi \circ \overline{\gamma})'(0) = \lambda (\varphi \circ \gamma)'(0)$, we set $\lambda [\gamma] := [\overline{\gamma}]$.

*Addition of tangent vectors:
Let $[\gamma_1], [\gamma_2] \in T_p M$, define a new curve $\hat{\gamma}$ by:
$t \mapsto \varphi^{-1}((\varphi \circ \gamma_1)(t) + (\varphi \circ \gamma_2)(t) - \varphi(p))$. Then since $(\varphi \circ \hat{\gamma})'(0) = (\varphi \circ \gamma_1)'(0) + (\varphi \circ \gamma_2)'(0)$ (check this!) we define: $[\gamma_1] + [\gamma_2] := [\hat{\gamma}]$. One can (and probably should) check that this vector space structure is independent of the chosen chart $(U,\varphi)$ and makes
$T_p M$ into a (real) vector space of dimension $n$.

