For $x\in\R^3$ and $X\in\so(3)$ put $u(t,x)=f(\exp(-tX)x)=\G(e^{tX})f(x)$: \begin{eqnarray*} \pa_tu(t,x) &=&\ftd t\G(e^{-tX})f(x)\\ &=&\lim_{s\to0}\frac{\G(e^{-tX-sX})f(x)-\G(e^{-tX})f(x)}s\\ &=&\lim_{s\to0}\frac{\G(e^{-sX})\G(e^{-tX})f(x)-\G(e^{-tX})f(x)}s\\ &=&\g(X)\G(e^{-tX})f(x) =\g(X)f(e^{-tX}x) \end{eqnarray*} and for $X=l_j$ we have by definition: $\g(X)=-iL_j$.
This is a particular case: Suppose $X$ is a complete vector field on a submanifold $M$ of $\R^d$ with flow $\theta$; for any smooth function $f:U\rar\R$ in an open neighborhood of $M$, the function $u:\R\times M\rar\R$, $P_tf(x)\colon=f(\theta(t,x))$ satisfies $\pa_tP_tf(x)=XP_tf(x)$, where $$ Xf(x)\colon=\sum_{j=1}^d\z_j(x)\pa_jf(x)~. $$