 Solution

Differentiate the circle group action at angle zero. The resulting real vector field isIn real coordinates $z=x+iy$, this is $K=-y{\partial }_x+x{\partial }_y$; the complex coefficients are conjugates, so it is indeed real. Its integral curves satisfy $\dot{z}=iz$ and hence $z(t)=e^{it}z(0)$, with period $2\pi$ away from the fixed points.

The affine coordinate does not cover the whole complex projective line, so check the other chart. With $w=1/z$, the same vector field is $K=-iw{\partial }_w+i\bar{w}{\partial }_{\bar{w}}$, smooth at $w=0$. It vanishes at $z=0$ and $w=0$, the two fixed poles. It therefore generates the stated global rotation of $S^2$, not merely a flow in one chart.