In Hamilton's Mathematical Gauge Theory he gives the following theorem:
Let $\varphi: G \rightarrow H$ be a homomorphism between Lie groups and $\varphi_*: \mathfrak{g} \rightarrow \mathfrak{h}$ the induced homomorphism on Lie algebras. Then $$\varphi(\exp{X}) = \exp{(\varphi_*X)} \quad \forall X \in \mathfrak{g}.$$
His proof is the following:
Consider the curve $\gamma(t) = \varphi(\exp{tx})$ for $t \in \mathbb{R}$. Then $$\gamma(0) = \varphi(e) = e$$ and $$\dot{\gamma}(t) = D_{(\exp{tX})} \varphi \Big( \frac{d}{ds}\Big|_{s=t} \exp{sX}\Big) \\ = D_{(\exp{tX})} \varphi (X_{(\exp{tX})}) \\ = D_e(\varphi \circ L_{(\exp{tX})})(X_e) \\ = D_e(L_{(\varphi(\exp{tX})}) \circ \varphi)(X_e) \\ = (\varphi_* X)_{\gamma(t)}.$$ Here we used that $\varphi$ is a homomorphism and $X$ left-invariant. We conclude that $\gamma$ is the (unique) integral curve of the left-invariant vector field $\varphi_*X$ through $e\in H$ and therefore $$\exp(\varphi_*X) = \gamma(1) = \varphi(\exp{X}).$$
In his notation, $e$ is the identity element of the Lie group, $D_x$ represents the differential evaluated at $x$, and $L_x$ represents left multiplication by $x$.
I am confused on how he what he does at each step on the string of equalities used in calculating $\dot{\gamma}(t)$. It looks like the first step he uses the chain rule. In the second step he evaluates the inner derivative and uses the fact that $X$ is left-invariant. What does he do for the remaining equalities?