Resolving varying $ω$ in Kepler's Law's Proof I'm having trouble understanding where $d^2r/dt^2$ comes from and what it stands for. What force is this? I'm not able to find any FBD's on google that mention any other force besides gravitational and centripetal force.
Also, why does $ω$ vary? Even if it is in an ellipse, and $ω$ is not constant, wouldn't $dr/dt$ also not be constant since the radius changes?
Here is the partial proof:
Kepler’s First Law
Note: I’m including the calculus derivation of the elliptic orbit, not to be found in the textbooks, just so you can see that it’s calculus, not magic, that gives this result.  This is an optional section, and will not appear on any exams.
We now back up to Kepler’s First Law: proof that the orbit is in fact an ellipse if the gravitational force is inverse square.  As usual, we begin with Newton’s Second Law: F = ma, in vector form. The force is $GMm/r^2$ in a radial inward direction. But what is the acceleration? Is it just $d^2r/dt^2$?  Well, no, because if the planet’s moving in a circular orbit it’s still accelerating inwards at $rw^2$ (same as v2/r) even though r is not changing at all. The total acceleration is the sum, so ma = F becomes:
$d^2r/dt^2-rw^2=-GM/r^2$
This isn’t ready to integrate yet, because w varies too. But since the angular momentum L is constant, $L = mr^2w$, we can get rid of w in the equation to give:
$d^2r/dt^2=-GM/r^2+r(L/mr^2)^2$
$=-GM/r^2+L^2/mr^3$
Here is the link to the website I found the proof on in case the screenshot isn't clear: http://galileo.phys.virginia.edu/classes/152.mf1i.spring02/KeplersLaws.htm
 A: What you are posting is just NEwton's second law.
The sum of forces must equal the product mass times acceleration.
The only force is the gravitational one, so "force of gravity = mass times acceleration"
$$-\dfrac{GMm}{r^2} = ma $$
Evidently, the mass of the planet cancels out, so you get
$$-\frac{GM}{r^2}=a$$
The only thing that is been doing here is that acceleration is a vector. Assuming it is a 2D movement (and it is, due to conservation of angular momentum), any vector can be split up in two components. Normally, we split accelration in $a_x$ and $a_y$. However, when dealing with orbits, it is more useful to use local coordinates: a tangential axis and a normal axis.
Both tangential and normal axis vary with time, since they move and rotate with the planet itself, but the key thing is that they are always orthogonal, so, Pithagoras still holds:
$$a^2 = a_t^2 +a_n^2$$
Then, recall that
$a_t=\frac{d^2r}{dt^2}$, while $a_n=-\omega^2 r$. Of course, $d^2r/dt^2$ stands for the second derivative of the radius' modulus repect to time, I guess you're familiarized with that.
So that's where your equiation comes from. What I don't see is why there is a missing square root, I think. In fact, we normally work with energy equations, rather than forces
If you use the energy equation...
$$E_c+E_p=E$$
$$ \frac{1}{2} m v^2 - \dfrac{GMm}{r}=E $$
$$\frac{1}{2}m \left[\left(\frac{dr}{dt}\right)^2+(\omega r)^2\right]- \dfrac{GMm}{r}=E $$
$$\frac{1}{2}m \left(\frac{dr}{dt}\right)^2+ \frac{1}{2}m\omega^2 r^2 - \dfrac{GMm}{r}=E $$
Finally, you also ask why $\omega$ varies. This is easily derived from conservation of angular momentum. Anyways, you couldn't have assumed it is constant, you should always consider it a variable, just in case, unless there is something that prevents it from varying. However, in this case it is clear: it does vary due to conservation of $L$.
