= Solution
Differentiate the <circle group> action at angle zero. The resulting real <vector field> is
$$
\boxed{K=iz\partial_z-i\bar z\partial_{\bar z}.}
$$
In 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. \b[It therefore generates the stated global rotation of $S^2$, not merely a flow in one chart.]
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