Sorry this going to be mildly long, I am having trouble obtaining what I think I should when I pull back the curvature form $S^3$ to $S^2$ via a local section, and would like to see if someone can point out a mistake in my reasoning. I also would prefer to avoid the identification of $S^2\cong \mathbb{C}P^1$.
Let $S^3$ be the unit three sphere, and $S^1$ the circle. There exists a free and proper right action on $S^3$ by $S^1$, given by: $$(z_1,z_2)\cdot e^{i\theta}=(z_1\cdot e^{i\theta},z_2\cdot e^{i\theta})$$ Since the action is free and proper we have that $S^3/S^1\cong S^2$ is a smooth manifold, and that the projection map $\pi:S^3\rightarrow S^2$ is a smooth submersion. By identifying $\mathbb{R}^2\cong \mathbb{C}$, we see that: $$S^2=\{(w,z)\in \mathbb C\times\mathbb R:|w|^2+z^2=1\}$$ Thus the map: $$\begin{align} F:S^3/S^1&\longrightarrow S^2\\ [z_1,z_2]&\longmapsto \left(2z_1\bar{z}_2, 2|z_1|^2-1\right) \end{align}$$ is a diffeomorphism as it is a well defined, smooth bijection, with an invertible differential map at each point. So the projection map $\pi:S^3\rightarrow S^2$ should be be given by: $$\pi((z_1,z_2))=(2z_1\bar{z}_2,2|z_1|^2-1)$$ as it is smooth, surjective, and satisfies $\pi((z_1,z_2)\cdot e^{i\theta})=\pi((z_1,z_2))$. I believe I can then write down a local section as $s:U=S^2\smallsetminus\{(0,1)\}\rightarrow S^3$ by: $$\begin{align} s(w,z)=\left(i\sqrt{\frac{1}{2}(1-z)}, \frac{1}{2}\frac{i\bar w}{\sqrt{\frac{1}{2}(1-z)}}\right) \end{align}$$ Which indeed takes image in $S^3$: $$\begin{align} \frac{1}{2}(1-z)+\frac{1}{2}\frac{|w|^2}{(1-z)}=&\frac{1}{2}(1-z)+\frac{1}{2}\frac{1-z^2}{(1-z)}\\ =&\frac{1}{2}(1-z)+\frac{1}{2}(1+z)\\ =&1 \end{align}$$ Furthermore: $$\begin{align} \pi\circ s(w,z)=&\left(2i\sqrt{\frac{1}{2}(1-z)}\cdot \frac{1}{2}\frac{-iw}{\sqrt{\frac{1}{2}(1-z)}}, \frac{|w|^2}{(1-z)}-1\right)\\ =&\left(w, \frac{1-z^2}{(1-z)}-1\right)\\ =&\left(w,1+z-1\right)\\ =&(w,z) \end{align}$$ so $\pi\circ s=\text{Id}_U$ and is thus a smooth section of $\pi$.
Finally, we have a connection one form $A$ on $S^3$ given by: $$A=(\bar{z}_1dz_1-z_1d\bar{z}_1+\bar{z}_2dz_2-z_2d\bar{z}_2)$$ Since $S^1$ is abelian, it follows that the curvature form: $$\begin{align} F^A=&dA+\frac{1}{2}[A,A]\\ =&dA\\ =&-(dz_1\wedge d\bar{z}_1+dz_1\wedge d\bar{z}_2) \end{align}$$
I want to pull this back $F^A$ back to $U$ by $s$, and this form on $U$ should extend to a globally defined two form on $S^2$ as $S^1$ is abelian. I have reason to believe that I should get the standard volume form of $S^2$ which is given by: $$ \omega=xdy\wedge dz+ydz\wedge dx+zdx\wedge dy $$ One can check with the angle parameterization of $S^2$: $$\phi^{-1}(\theta,\phi)=(\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta)$$ that: $$(\phi^{-1})^*\omega=\sin\theta d\theta\wedge d\phi$$ so this is indeed the volume form on $S^2$.
Proceeding, I calculate: $$\begin{align} s^*(dz_1)=&d(z_1\circ s)\\ =&\frac{-idz}{2\sqrt{2}\sqrt{1-z}} \end{align}$$ while: $$d(\bar{z}_1\circ s)=\frac{idz}{2\sqrt{2}\sqrt{1-z}}$$ hence: $$s^*(dz_1\wedge d\bar{z}_1)=d(z_1\circ s)\wedge d(\bar{z}_1\circ s)=0$$ Furthermore: $$\begin{align} d(z_2\circ s)=\frac{i\bar{w}dz}{2\sqrt{2}(1-z)^{3/2}}+\frac{id\bar{w}}{\sqrt{2}\sqrt{1-z}} \end{align}$$ and: $$\begin{align} d(\bar{z}_2\circ s)=\frac{-i{w}dz}{2\sqrt{2}(1-z)^{3/2}}+\frac{-id{w}}{\sqrt{2}\sqrt{1-z}} \end{align}$$ so: $$s^*(dz_2\wedge d\bar{z}_2)=\frac{\bar w}{4(1-z)^2}dz\wedge dw +\frac{w}{4(1-z)^2} d\bar w\wedge dy +\frac{1}{2(1-z)}d\bar{w}\wedge dw$$
Letting $w=x+iy$ we obtain that: $$s^*(dz_2\wedge d\bar{z}_2)=\frac{-ix}{2(1-z)^2}dy\wedge dz+\frac{-iy}{2(1-z)^2}dz \wedge dx+\frac{-i}{(1-z)}dy\wedge dx$$ Hence: $$s^*F^A=\frac{ix}{2(1-z)^2}dy\wedge dz+\frac{iy}{2(1-z)^2}dz \wedge dx+\frac{i}{(1-z)}dy\wedge dx$$ Which to me does not at all seem like it is equal to $\omega$. I know that it should be some scalar times $\omega$ since the book I'm using shows that the volume form in a chart on $S^2$ gives rise to the curvature form. I could do something similar, but I wanted to start with the curvature form and go backwards since the other way seems contrived. Any help would be greatly appreciated.
Edit to Add Integral:
We pull $s^*F$ back by the angle parameterization of $S^2$: $$\phi^{-1}(\theta,\phi)=(\sin\theta\cos\phi,\sin\theta\sin\phi, \cos\theta)$$ which leaves out a set of measure zero in $S^2$, hence:
$$\int_{S^2}s^*F=\int_0^\pi\int_{0}^{2\pi}\phi^{-1*}(s^*F)$$
We see that: $$ \begin{align} \phi^{-1*}(s^*F)=\frac{i\sin\theta}{1-\cos\theta}d\theta\wedge d\phi \end{align} $$
So: $$ \begin{align} \int_{S^2}s^*F=&2\pi \int_0^\pi\frac{\sin\theta}{1-\cos\theta}\\ \end{align} $$ which clearly doesn't converge so something is off here...
Edit to Add:
Completely messed up the calculation of the pull back. It ends ups being
$$\phi^{-1*}(s^*F)=\frac{i}{2}\sin\theta d\theta\wedge d\phi$$
So that makes much more sense.