Let $M$ be a manifold of dimension $d$, and $p$ some fixed point on $M$. Define the tangent space $$T_p(M):=\{v_{\gamma,p}:\gamma\text{ is a curve in $M$ passing through $p$}\}.$$ We want to construct a basis for the tangent space which we know can be made into a vector space. To do this, we choose curves (which pass through $p$), $\gamma_{(a)}:\mathbb{R}\to U$, with $a\in[1,d]$, where $U$ is a neighborhood about $p$, such that $x^b\circ\gamma_{(a)}(\lambda)=\delta^b_a\cdot\lambda$, where we have chosen a chart $x$ on $U$ so $x(p)=\mathbf{0}\in\mathbb{R}^d$. Moreover, we assume wlog that $\gamma_{(a)}(0)=p$. This can be arranged by making suitable translations in domain. Now we set $e_a=v_{\gamma_{(a)},p}(f)=(f\circ\gamma_{(a)})'|_0$. We claim $e_a$ forms a basis for $T_p(M)$.
To see this, we choose an arbitrary function $f:M\to\mathbb{R}$. We have $e_a(f)=(f\circ\gamma_{(a)})'|_0=(f\circ x^{-1})'|_{x(\gamma_{(a)}(0))}\cdot(x\circ\gamma_{(a)})'|_0$. Now, the latter term, by choice of $x,\gamma_{(a)}$ is $\delta^b_a$. The former is $\partial_b(f\circ x^{-1})|_{x(p)}$. So overall, this evaluates to $\partial_a(f\circ x^{-1})$. It's easy to see the decomposition of any element of the tangent space can be constructed using these guys.
However, to me, this seems like a very messy construction. I can understand the geometric intuition behind the choices of $\gamma_{(a)}, x$ and what-not, but I struggle to see why such $\gamma_{(a)}$ must exist. How can we be guaranteed such curves exist?
I suppose a naive way to see this is to choose $\gamma_{(a)}$ to be the curve defined by the action $x^{-1}$ on the $a$th axis in $\mathbb{R}^d$ passing through $0$. But I'm having trouble formalizing this. Any help would be appreciated.
First, let us fix some notations. If $v=(v_1, \ldots, v_n)\in\mathbb R^n$ define
$$|v|=(v_1^2+\ldots+v_n^2)^{1/2}.$$ For $a\in \mathbb R^n$ and $r>0$, write
$$B(a, r)=\{x\in\mathbb R^n: |x-a|<r\}.$$
Take a local chart $(U, \phi)$ around $p$. Here $\phi: U\rightarrow \phi(U)$ is a homeomorphism between $U$ and the open set $\phi(U)\subset \mathbb R^n$.
Since $\phi(p)\in \phi(U)$ and $\phi(U)$ is open, there is $r>0$ such that $B(\phi(p), r)\subset \phi(U)$. Fix any $v\in \mathbb R^n$ and take $\varepsilon=r/(1+|v|)$. Then,
$$|\phi(p)+tv-\phi(p)|=|tv|=|t||v|<\frac{r}{1+|v|}|v|<r$$
and therefore $\phi(p)+tv\in B(\phi(p), r)\subset \phi(U)$ for every $t\in (-\varepsilon, \varepsilon)$. Define $\gamma: (-\varepsilon, \varepsilon)\rightarrow \phi(U)$ setting
$$\gamma(t)=\phi(p)+tv.$$
Then $\phi^{-1}\circ \gamma: (-\varepsilon, \varepsilon)\rightarrow U$ is such that
$$\phi^{-1}(\gamma(0))=\phi^{-1}(\phi(p))=p.$$