Suppose that $M$ is a Riemannian manifold.
This proposition
For each $p \in M$ there exist an open set $\mathcal U \subset TU$, where $(U,\mathbf x)$ is a system of coordintes at $p$ and $(p,0) \in \mathcal U$, a number $\delta > 0$ and a $C^\infty$ map $\varphi : (-\delta,\delta) \times \mathcal U \to TU$, such that $t \mapsto \varphi(t,q,v)$ is the unique trajectory of the vector field $G$ (the geodesic field on $TM$) which satisfies the initial condition $\varphi(0,q,v)=(q,v)$ for each $(q,v) \in \mathcal U$.
Let $\gamma := \pi \circ \varphi$, where $\pi : TM \to M$ is the canonical projection defined by $\pi(q,v) = q$. Also choose in particular $$ \mathcal U := \{(q,v) \mid q \in V, v \in T_qM, |v| < \epsilon_1\} $$
These would imply this proposition:
Given $p \in M$, then there exist an open set $V \subset M$, $p \in V$, numbers $\delta > 0$ and $\epsilon_1 > 0$, and a $C^\infty$ map $\gamma : (-\delta,\delta) \times \mathcal U \to M$, where $$ \mathcal U := \{(q,v) \mid q \in V, v \in T_qM, |v| < \epsilon_1\} $$ such that the curve $t \mapsto \gamma(t,q,v)$ for all $t \in (-\delta,\delta)$, is the unique geodesic of $M$ which, at the instant $t=0$, passes through $q$ with velocity $v$ for each $q \in V$ and for each $v \in T_qM$ with $|v| < \epsilon_1$.
And finally this lemma (homogeneity of a geodesic):
If the geodesic $\gamma(t,q,v)$ is defined on the interval $(-\delta,\delta)$, then the geodesic $\gamma(t,q,av)$, where $a > 0$, is defined on the interval $(-\frac{\delta}a,\frac{\delta}a)$ and $$\gamma(t,q,av)=\gamma(at,q,v).$$
The proof of the lemma begins by introducing a curve $h : (-\frac{\delta}a,\frac{\delta}a) \to M$. Then this curve satisfies $h(0)=q$ and $\frac{dh}{dt}(0)=av$.
I got that $$ h(0)=\pi(\varphi(a \cdot 0,q,v))=\pi(\varphi(0,q,v))=\pi(q,v)=q, $$ but how do I justify $\frac{dh}{dt}(0)=av$? I tried the following: $$ h'(0)=\pi'(\varphi(a \cdot 0,q,v))\varphi(a \cdot 0,q,v)=\cdots \, ?? $$
For reference, this discussion came from pages 63-64 of Riemannian Geometry by Manfredo do Carmo.