A different way of calculating the surface area of the sphere I just don't understand this. I know $A_{sphere} = 4\pi r^2$ and the circumference $C_{sphere}  = 2\pi r$, so why can't I just sum up (integrate) all the circumferences to get the area? That is, why is
$$
\int_0^\pi C_{sphere} \ d\theta = \int_0^\pi 2\pi r\ d\theta = 2\pi^2 r \neq 2\pi r^2 = A_{sphere}
$$
I know I'm missing a core concept here, but it seems completely fine to take the integral of all the circumferences (i.e. sum of the infinitesimally thin circumferences) in the sphere to get the area.

What you denote by $C_{sphere}$ should be called the circumference of a circle. Imagine that you stuck together many circles to form a cylinder of length $\pi$. Then
$$\int_{0}^{\pi} C_{sphere} \, d\theta$$
is the surface area of this cylinder (not counting the two disks at both ends of the cylinder). There is no reason why this should be equal to the surface area of a sphere!
The problem is that there is no way to slice the sphere to get "$\pi$" circles of radius $r$. If you try slicing an apple into many parallel slices, you will get circles of different sizes (most of them having radius less than $r$). Assume, you are slicing perpendicular to the stem and $\theta$ is the angle between the stem and the point you start a particular slice. Then you will get a circle of radius $r \sin \theta$. The length (or circumference) of this slice is therefore
$$2 \pi \cdot r \sin \theta.$$
If you measure how wide the skin is on this slice (not the slice itself!), you will get
$$r \, d\theta.$$
From this you can get the total area by integrating
$$\int_{0}^{\pi} 2 \pi r^2 \sin \theta \, d\theta = 4 \pi r^2.$$

