From Lee's book regarding smooth manifolds. He has the following proposition
Suppose $M$ is a smooth manifold with or without boundary and $p \in M$. Every $v \in T_pM$ is the velocity of some smooth curve in $M$
He then gives the following proof for the case without boundary.
First suppose that $p \in \operatorname{Int} M$ (which includes the case $\partial M = \emptyset $). Let $(U, \varphi)$ be a smooth coordinate chart centered at $p$, and write $v= \sum_{i}v^i \frac{\partial}{\partial x^i} \bigg|_{p}$ the coordinate basis. For sufficiently small $\varepsilon >0 $, let $\gamma :(-\varepsilon, \varepsilon) \to U$ be the curve whose coordinate representation is $$\gamma(t)= (tv^1, \dots,tv^n)$$ (Remember, this really means $\gamma(t)=\varphi^{-1}(tv^1, \dots tv^n)$.) This is a smooth curve with $\gamma(0)=p$, and the computation above shows that $\gamma'(0)=\sum_{i}v^i \frac{\partial}{\partial x^i} \bigg|_{\gamma(0)}=v$.
Few questions. I don't quite understand the note he gave that
Remember, this really means $\gamma(t)=\varphi^{-1}(tv^1, \dots tv^n)$.
the map $\gamma$ is defined from some interval to the manifold $M$, but why would this interval be contained in the space say $\Bbb R^n$ that's the chart's codomain? How does it make sense to say that $\gamma$ would be the inverse of $\varphi$?
Secondly how does he get the result
and the computation above shows that $\gamma'(0)=\sum_{i}v^i \frac{\partial}{\partial x^i} \bigg|_{\gamma(0)}=v$.
what computation does he mean? The way he defines $\gamma'$ is that $$\gamma'(0)=d\gamma \left(\frac{d}{dt} \bigg|_{t_0} \right) \in T_{\gamma(t_0)}M?$$