Let $M$ be a smooth manifold.
A tangent vector of $M$ at $p$ is an equivalence class $[\gamma]$ of smooth curves $\gamma: (-\epsilon, \epsilon) \to M$ with $\gamma(0) =p$, where $\gamma_1 \sim \gamma_2$ means that there is a chart $\phi$ around $p$ such that $(\phi\circ \gamma_1)'(0) = (\phi\circ \gamma_2)'(0)$.
Question: Is $\epsilon > 0$ fixed? Or is possible that $\gamma_1 \sim \gamma_2$ where the two curves have a different domain? I guess it's the latter, but wanted to be sure.
The intuitive picture of "curves having the same velocity at a given point" is pretty clear, but, as you point out, there are some technical obstacles to making it precise. There are a number of equivalent constructions. Here's one recipe (among many others):
Choose $p\in M$.
Choose a chart $\varphi:U\to \mathbb{R}^m$ containing $p$.
Consider the set of curves $\gamma:I\to M$ where $I\subseteq\mathbb{R}$ is an interval containing $0$ and $\gamma(0)=p$.
Define an equivalence as follows: Given $\gamma_1:I_1\to M$, and $\gamma_2:I_2\to M$ as above, say $\gamma_1\sim\gamma_2$ if there are restrictions of domain to open intervals $J_1\subseteq I_1$, $J_2\subseteq I_2$ containing $0$ such that $\gamma_1|_{J_1}$ and $\gamma_1|_{J_2}$ both map into $U$ and $\left.\frac{d}{dt}\varphi(\gamma_1|_{J_2}(t))\right|_{t=0}=\left.\frac{d}{dt}\varphi(\gamma_2|_{J_2}(t))\right|_{t=0}$ in the vector calculus sense.
From then, it needs to be checked that these restrictions always exist, and that the equivalence is independent of the choice of restriction, and that the equivalence classes are independent of the choice of chart.