While researching the formulas for these calculations, I am completely stumped by how the ones circled in red are derived. I think the surface area formula is wrong here since that's the formula for the surface area of the label of a cylinder.
Does anyone know the correct formula, or if that is the correct formula, why it is the correct formula?
Both can be derived using some basic integral calculus. Take into account that the equation of a circle of radius r is $y=\sqrt{r^2-x^2}$, thus $y'=\frac{x}{\sqrt{r^2-x^2}}$. So \begin{align*} V_s&=\int_{r-h}^r\pi(\sqrt{r^2-x^2})^2dx=\pi\int_{r-h}^rr^2-x^2dx\\ A_s&=\int_{r-h}^r2\pi\sqrt{r^2-x^2}\sqrt{1+\left(\frac{x}{\sqrt{r^2-x^2}}\right)^2}dx \end{align*} To see how these formulas are derived you can check any clasical calculus book as could be Stewart's or Thomas's. Feel free to reply if you need help when computing any of those integrals.
Now, if you want to avoid Calculus, use Cavalieri's principle, i will ilustrate how to use it to proof the volume of half the sphere, a restriction to my argument shall yield the formula you are looking for:
If you consider a sphere of radius $r$ when you consider the transversal section a distance $s$ away from the center it has area $A_T=\pi r^2-\pi s^2$; which can be seen as the area of a circle of radius $r$ minus the area of a circle with radius $s$. Now consider a cilinder whose base's radius is $r$ and with height $r$ without a cone of base $r$ and height $r$ as in the image.
By Cavalieri's principle this shape has the same volume as half sphere and it is clearly $$V=\pi(r^2)r-\frac{1}{3}\pi r^2r=\frac{2}{3}r^3$$ To deal with your problem consider calculating only a part of the volume of the image instead of all.
I hope it is useful