I understand the proof, up until the very last (most important!) part. The set-up is this: if we have a smooth right action $(p,g)\mapsto p\cdot g$ of a Lie group $G$ on a smooth manifold $M$ then each $X\in \text{Lie} (G)$ induces a smooth global flow on $M$ via $\theta(t,p): (t,p)\mapsto p\cdot \exp tX.$ Let $\hat X$ be the infinitesimal generator of this flow, so that $\hat X_p=\theta'(0,p).$ Finally, define $\hat \theta $ to be the map that sends $X$ to $\hat X.\ \hat \theta $ is called the infinitesimal generator of the action.
The theorem is a converse. It says that if $\hat \theta:\text{Lie}(G)\to \mathfrak X(M)$ is a homomorphism such that $\hat \theta (X)$ is complete for every $X\in \text{Lie}(G)$ then there is a unique smooth right $G$-action on $M$ whose infinitesimal generator is $\hat \theta.$
The proof is constructive, using covering spaces and foliations to produce the action. Set $\hat \theta (X)=\hat X$ and let $\eta_{\hat X}$ be the flow of $\hat X$. The conclusion of the theorem is that there is an action as advertised, given by $p\cdot g=\eta_{\hat X}(1,p)$ for $g=\exp X$ in a neighborhood of $e$ and extended to all of $G$.
The claim is therefore that $\hat \theta$ is the infinitesimal generator of the action.
I have no problem understanding the construction, but I do not see how it "works". Lee says it's an "immediate consequence" of a particular line in the proof, namely
I tried to calculate this directly, without success. I need to show that the infinitesimal generator of the flow $\Theta(t,p)=p\cdot \exp tX$ is $\hat X$, and the above implies that $\Theta(t,p)=\eta_{t\hat X}(1,p)$, which is an integral curve of $t\hat X$ so so is $\Theta_1(t,p)=\eta_{(t-1)\hat X}(0,p)$ and $\Theta_1'(0,p)=(0-1)\hat X_p=-\hat X.$ I think I am missing something more fundamental. Can someone point me in the right direction?
edit: here is the answer:
It suffices to show that $\eta_{t\hat X}(1,p)=\eta_{\hat X}(t,p)$ for all $t\in \mathbb R.$ We will appeal to the uniqueness of integral curves through $p\in G.$
$\gamma:\mathbb R\to G$ defined by $\gamma(s)=\eta_{t\hat X}(s,p)$ is an intgral curve of $t\hat X$ starting at $p$.
Consider $\sigma:s\mapsto \eta_{\hat X}(ts,p).$ Then, $\sigma(0)=p$ and $\frac{d\sigma}{ds}=\frac{td(\eta_{\hat X}(ts,p))}{ds}=t\hat X_{\eta_{\hat X}(ts,p)}=t\hat X_{\sigma(s)}$, which means that $\sigma$ is an integral curve of $t\hat X$ starting at $p.$
The conclusion is then that $\eta_{t\hat X}(s,p)=\eta_{\hat X}(ts,p)$ and setting $s=1$ finishes the proof.