I want to check that I understand this definition correctly and that my example makes sense.
Let $\varphi:G\to Diff(M)$ be a action of a Lie group $G$ on a smooth manifold $M$. We say that $\varphi$ is a smooth action if the evaluation map, i.e. $$ev:G\times M\to M$$ given by $ev(g,m)=\varphi(g)(m)$, is a smooth map.
Example:
Let's show that action $\varphi$ of $S^1$ on $\mathbb{R}^2$, given by $\varphi:S^1\to Diff(\mathbb{R}^2)$, $\varphi(\theta)(x,y)=(x\cos(\theta)-y\sin(\theta),x\sin(\theta)+y\cos(\theta))$, is a smooth action. It's enough to show that the evaluation map $$ev:S^1\times \mathbb{R}^2\to \mathbb{R}^2\text{ is smooth}.$$ Indeed, consider the open subset $U=S^2-\{N\}$, a homeomorphism $\psi:U\to\mathbb{R}$ with $\psi(x,y)=\frac{x}{1-y}$, and its inverse $\psi^{-1}:\mathbb{R}\to U$ with $\psi^{-1}(t)=\left(\frac{2t}{1+t^2},\frac{t^2-1}{t^2+1}\right)$.
Then we have a map $$(ev\circ\psi^{-1}(t))(t,(x,y))=ev\left(\left(\frac{2t}{1+t^2},\frac{t^2-1}{t^2+1}\right),(x,y)\right)$$$$=\left(x\frac{2t}{1+t^2}-y\frac{t^2-1}{t^2+1},x\frac{t^2-1}{t^2+1}+y\frac{2t}{1+t^2}\right)$$ which is obviously a smooth map. A similar map we can obtain by considering another open subset of $S^1$. Therefore, the action $\varphi$ is smooth.