Let M be a smooth manifold. Let $ J=[0,1] $ If $ \gamma : J \to M $ is a smooth curve, then for each $ t \in J $, the velocity vector $ \gamma '(t) \in T_{\gamma(t)} M$
I am self studying smooth manifolds and came across this statement. I am aware that each tangent vector in the tangent space of a manifold can be identified with a derivation; namely the directional derivative at a particular $ p \in M $
I cannot see how $ \gamma ' (t) $ can be written using the basis vectors of the tangent space
One possible definition for $\gamma'$ is $\gamma = {d\gamma} \colon T(a,b) \to TM$, $d\gamma(v)(f) = v(f\gamma)$.
However, we may identify $T(a,b)$ with $(a,b)$ as $t \mapsto \frac{d}{dx}|_t$ and define $\gamma'(t) = d\gamma\left(\frac{d}{dx}|_{x=t}\right)$.
By definition, this is a derivation on $T_{\gamma(t)}M$ that sends a smooth function $f \colon M \to \mathbb{R}$ to
$$ \frac{d}{dx}|_{x=t}(f\gamma) = (f\gamma)'(t). $$
In particular, if $\varphi$ is a chart of $M$ around $\gamma(t)$, there are unique $b_1,\ldots,b_n \in \mathbb{R}$ such that $\gamma(t)' = \sum_i b_i \frac{\partial}{\partial \varphi^i}|_{\gamma(t)}$.
Moreover, evaluating at each coordinate function $\varphi^i \colon M \to \mathbb R$ we see that
$$ b_i = \gamma'(t)(b_i) = (\varphi^i \gamma)'(t) $$
and thus
$$ \gamma'(t) = \sum_{i=1}^n (\varphi^i \gamma)'(t) \frac{\partial}{\partial \varphi^i}\Big|_{\gamma(t)}. $$