Theorem 2.2. If $X$ is a $C^\infty$ vector field on the open set $V$ in the manifold $M$ and $p \in V$ then there exist an open set $V_0 \subset V$ , $p \in V_0$, a number $\delta > 0$, and a $C^\infty$ mapping $\varphi : (-\delta,\delta) \times V_0 \to V$ such that the curve $t \to \varphi(t,q)$, $t\in(-\delta,\delta)$, is the unique trajectory of $X$ which at the instant $t_0$ passes through the point $q$, for every $q \in V_0$.
The theorem above is a result from differential equations. There's the following applications to the geodesic field (which I think is more of an adaptation).
For each $p \in M$ there exist an open set $\mathcal{U}$ in $TU$, where $(U,x)$ is a system of coordinates at $p$ and $(p,0) \in \mathcal{U}$, a number $\delta > 0$ and a $C^{\infty}$ mapping, $\varphi : (-\delta,\delta) \times \mathcal{U} \to TU$ such that $t \to \varphi(t,q,v)$ is the unique trajectory of $G$ which satisfies the initial conditions $\varphi(0,t,v) = (q,v)$ for each $(q,v) \in \mathcal{U}$.
Right afterwords
It is possible to choose $\mathcal{U}$ in the form $$\mathcal{U} = \left\{ (q,v) \in TU ; q \in V\ \ \text{and}\ \ v \in T_q M \text{ with } \;\; |v| < \epsilon_1 \right\}$$ where $V \subset U$ is a neighborhood of $p \in M$.
The question is : why can choose the $\mathcal{U}$ in that specific form? Also, given the notation $|v| < \epsilon_1$ shouldn't we assume that $M$ is a Riemannian manifold?
Sets of the form $\mathcal U$ form a neighborhood basis of the point $(p,0) \in TU$, so you are free to choose $\mathcal U$ of that form.
And regarding the notation $|v| < \epsilon_1$, what's being used here is not the full panoply of Riemannian geometry, but just an arbitrary smoothly varying norm on the tangent spaces at points of $V$, chosen for the very purpose of expressing a neighborhood basis of $(p,0)$.