Fix $p\in M$. Consider all smooth curves $c : (-\epsilon,\epsilon) \rightarrow M$ with $c(0) = p$. We say that two such curves $c_1$ and $c_2$ are equivalent if there exist some smooth chart $(U,\phi)$ with $p \in U$, such that $$\frac{d}{dt}\phi \circ c_1(0) = \frac{d}{dt}\phi \circ c_2(0).$$I am trying to show that if this is true for some chart, then it must be for every other chart that contains $p$.
Attempt: Assume the condition holds for $(U,\phi)$. Pick another chart $(V,\psi)$. Then $\psi \circ \phi^{-1} = \varphi \Leftrightarrow \psi = \varphi \circ \phi$ is smooth on $\psi(U \cap V)$. Thus, \begin{align*} \frac{d(\psi \circ c_1)}{dt}(0) & = \frac{d((\varphi \circ \phi) \circ c_1)}{dt}(0)\\ & = D_{\phi(p)}\varphi \circ D_0(\phi \circ c_1)\\ & = D_{\phi(p)}\varphi \circ D_{0}(\phi \circ c_2)\\ & = \frac{d(\varphi \circ (\phi \circ c_2))}{dt}(0)\\ &= \frac{d(\psi \circ c_2)}{dt}(0). \end{align*}