Here is the result I want to prove
Suppose the Killing field $X$ is 0 in a certain $x_0$. Show that if $(\varphi_t)_{t\in\mathbb{R}}$ is the flow of $X$, and if $Y\in T_{x_0}M$, then$$\varphi_t(\text{exp}_{x_0}(sY)) = \text{exp}_{x_0}(s(\varphi_t)_*(Y ))$$
This is my proof:
$$\varphi_t(\text{exp}_{x_0}(sY)) = \varphi_t(\gamma_1(s)) = (\varphi_t\circ\gamma_1)(s)$$ where $\gamma_1$ is the unique geodesic such that $\gamma_1(0) = x_0$ and $\dot{\gamma}_1(0) = Y$. So this quantity is the position of the particule at time $s$ following the geodesic $\varphi_t\circ\gamma_1$. $$\text{exp}_{x_0}(s(\varphi_t)_*(Y )) = \gamma_2(s)$$ where $\gamma_2$ is the unique geodesic such that $\gamma_2(0) = x_0$ and $\dot{\gamma}_2(0) = (\varphi_t)_*Y$. So this quantity is the position of the particule at time $s$ following the geodesic $\gamma_2$.
So if $\varphi_t\circ\gamma_1$ is such that $(\varphi_t\circ\gamma_1)(0) = x_0$ and $\dot{(\varphi_t\circ\gamma_1)}(0) = (\varphi_t)_*Y$ then we have our result by unicity of geodesics. $$\dot{(\varphi_t\circ\gamma_1)}(0) = (\varphi_t)_*\dot{\gamma}_1(0) = (\varphi_t)_*Y$$
But for the first condition, I'm not sure $$(\varphi_t\circ\gamma_1)(0) = \varphi_t(x_0) = x_0$$ Why is the last true? Since the field is $0$ in $x_0$ intuitively, it is clear because a particule at position $x_0$ will never leave $x_0$.