I'm a little confused by this notation:
Solution. The sphere $S^2$ has a parametrization by spherical coordinates (Figure 23.3(b)): $$F(\varphi,\theta)=(\sin\varphi\cos\theta,\sin\varphi\sin\theta,\cos\varphi)$$ on $D=\{(\varphi,\theta)\in\mathbb{R}^2\mid0\leq\varphi\leq\pi,0\leq\theta\leq2\pi\}$. Since $$F^*x=\sin\varphi\cos\theta,\quad F^*y=\sin\varphi\sin\theta,\quad\text{and }\,\, F^*z=\cos\varphi,$$
What is meant by $F^\ast x$? Previously the definitions of pullbacks of linear maps and forms were introduced, but what exactly is this $F^\ast x$?
If the form being considered is $\frac{dy\wedge dz}{x}$, then
$$F^*\omega=\dfrac{F^* dy\wedge F^* dz}{F^*x}=\sin\varphi\,d\varphi\wedge d\theta.$$
Is it a property of pullbacks that $F^\ast(g\omega)=F^\ast(g) F^\ast \omega$? I didn't find this in the book. What if I had $\frac{dy\wedge dz}{\sqrt{x^2+z^2}}$?
$x,y$, and $z$ denote the three coordinates in $\mathbb R^3$. So, $F^*x$ means that you are pulling the first coordinate back to spherical coordinates; in other words, $F^*x = x(F(\phi,\theta))$ which outputs the first coordinate of $F$.
As for your question on $F^*(g\omega) = F^*(g)F^*(\omega)$. I am assuming that $g$ is a scalar function and $\omega$ is a $k$-form in $\mathbb R^3$; in this case, yes, the equation holds true.
As for the final concern, note that $F^*(d\omega) = d(F^*\omega)$, we have $$\begin{split}F^*(\frac{dy\wedge dz}{\sqrt{x^2 + z^2}}) &= F^*(\frac{1}{\sqrt{x^2 + z^2}})F^*(dy\wedge dz)\\ &= \frac{1}{\sqrt{(\sin\phi \cos\theta)^2 + \cos^2\phi}} F^*(dy\wedge dz) \end{split}$$ Now, for $F^*(dy\wedge dz)$, one way to evaluate it is by pulling back each differential term then wedging the results as follows: $$\begin{split}F^*(dy\wedge dz)&= F^*(dy)\wedge F^*(dz)\\ &= d(F^*y) \wedge d(F^*z)\\ &= d(\sin\phi\sin\theta) \wedge d(\cos\phi)\\ &= (\cos\phi\sin\theta \ d\phi + \sin\phi\cos\theta \ d\theta) \wedge (-\sin\phi \ d\phi)\\ &= -\cos\phi\sin\theta \sin\phi \ d\phi \wedge d\phi - \sin^2\phi\cos\theta \ d\theta \wedge d\phi\\ &= 0 + \sin^2\phi\cos\theta \ d\phi \wedge d\theta \end{split}$$
We end up with $$F^*(\frac{dy\wedge dz}{\sqrt{x^2 + z^2}}) = \frac{\sin^2\phi\cos\theta}{\sqrt{(\sin\phi \cos\theta)^2 + \cos^2\phi}}d\phi \wedge d\theta $$