In Proposition 1.7.12 in Hamilton's Mathematical Gauge Theory he states
Let $G$ be a Lie group and $X$ a left-invariant vector field. Then its flow $\phi_t(p)$ through a point $p \in G$ is defined for all $t \in \mathbb{R}$, $$\phi:\mathbb{R} \times G \rightarrow G\\ (t,p) \mapsto \phi_t(p),$$ and given by $$\phi_t(p) = p \cdot \exp tX = R_{\exp tX}(p) = L_p (\exp tX).$$
In his notation $L_x$ and $R_x$ are left and right multiplication by $x$, and $D_e$ is the differential evaluated at $e$. His proof is
Define $\phi_t(p)$ for all $t \in \mathbb{R}$ by the right-hand side. It is clear that $$\phi_0(p) = p \cdot \exp(0) = p.$$ Furthermore, $$\frac{d}{dt}\Big|_{t = s} \phi_t(p) = \frac{d}{d\tau} \Big|_{\tau = 0} L_p(\exp sX \cdot \exp \tau X) \\ = D_e L_{p \exp sX}(X_e) \\ = X_{p \exp s X} \\ = X_{\phi_s(p)},$$ since $X$ is left-invariant. This implies the claim by uniqueness of solutions of ordinary differential equations.
I am having some trouble understanding what he did in the first two lines/equalities. The first seems to be a change of variables. The second appears to be using the chain rule, but I am not seeing how he obtains $X_e$ or why he is evaluating the differential at $e$.
The first line is the change of variable $t = s + \tau$, using the fact that $$\exp tX = \exp ((s+\tau)X) = \exp(sX + \tau X) = \exp sX \cdot \exp \tau X.$$ The second line is the associativity of the product, and then the chain rule. First, one has $$ L_p (\exp sX \cdot \exp \tau X) = p \cdot (\exp sX \cdot \exp \tau X) = (p\cdot \exp sX)\cdot \exp \tau X = L_{p\cdot \exp sX}(\exp\tau X). $$ Applying the chain rule, one has that \begin{align} \frac{d}{d \tau} L_p (\exp sX \cdot \exp \tau X) &= \frac{d}{d\tau} L_{p\cdot \exp sX}(\exp\tau X) \\ &= \left(D_{\exp \tau X}L_{p\cdot \exp sX}\right)\left(\frac{d}{d\tau}\exp\tau X\right), \end{align} and one concludes by evaluating at $\tau = 0$, since $\exp (0X) = \exp 0 = e$ and $(\frac{d}{d\tau}\exp \tau X)|_{\tau = 0} = X_e$.