Why is electrostatic force a function of the product of charges, not the sum? When you have two forces , you ADD them to get the total. If both of us push on a car, the total force on the car is the sum of our forces. But, If we have an electron approaching another electron, the force between them is given by Coulomb's law:
f=k q1 * q2 / r^2
How does one derive this formula from first principles? And why is it not the sum of q1 and q2?
The first flaw in this argument is that the electrons have a spherical electric field in 3d space. Their sum would only be valid in one dimensional space.
Consider two electrons in Abbot's world named 'Flatland' for simplicity's sake. Now imagine that the electric field of each electron is a circle.
 electron 
Imagine a force vector,F which we can break down into x and y components like so:
vectors X and Y 
Fx = cos(A) * F
We can calculate the Total force to the right by integrating from -pi/2 to pi/2 like so:
total force to the right
= F*2
Now imagine another electron to the right. It too has a bunch of force vectors whose x component we can calculate
All the horizontal components that will interact with another electron to the right will be the sum of all Fx so we integrate from -pi/2 to pi/2. Its total will also be F*2
So the sum of all forces pushing the electrons apart is F *4.
or Q * 4
not Q ^ 2
I could extend this argument to 3 space but it still does not give Q ^ 2.
 A: 
why is electrostatic force a function of the product of charges, not the sum?

Stipulate that $q_2$ is zero, i.e., that particle 2 is electrically neutral (uncharged).
If the electric force between the two particles is proportional to the product of their charges, then the electric force between an electrically charged particle and an electrically uncharged particle is zero.
However, if the electric force between the two particles is proportional to the sum, there would be an electric force between an electrically charged particle and an electrically uncharged particle.
Which model better fits what is actually observed?

UPDATE:
From a comment by the OP:

I am looking for a fundamental explanation of WHY we should use the
  product of charges in the numerator of Coulomb’s law

It still isn't clear to me what you're looking for.  In your question, you open with an example of adding forces together to get the total force and then use this to question why the product of the charges is used in the Coulomb force law.  But this seems like a category error to me.
Consider the case that there are three charges.  The total electric force on one of the charges is just the sum of the Coulomb force due to each of the other charges.  Similarly, and in reference to your opening paragraph, the total force on the car is the sum of each individual's pushing force.
On my view, the use the car example as a legitimate motivation for seeking the "fundamental explanation" for the product in Coulomb's law is just plain confused.

From another comment by the OP:

Yeah I guess your logic clearly works. But I might counter with the
  argument that there really is no uncharged particle.

Even if one were willing to entertain that objection, one can stipulate that $q_1 \gg q_2$ and then you have the result that the electric force between the two charges is (effectively) independent of $q_2$.
Again, which model better fits what is actually observed?
A: We can possibly apply this logic.
Due to superposition, the force the first charge applies to the second charge should be proportional to the value of the first charge, i.e., doubling the value of the first charge should double the force, etc.
For the same reason, the same should be true for the force the second charge applies to the first.
Since it is essentially the same force (third Newton's law), it has to be proportional to the values both charges.
A: A lone point charge, $q_1$, makes an electric field with a magnitude that is proportional to the charge (at fixed $r$):
$$ {\vec E} = \frac {q_1} {4\pi\epsilon_0}\frac{\hat r}{r^2}$$
The force on a charge, $q_2$, in an electric field is proportional to both the charge and the field:
$$ {\vec F} = q_2{\vec E} $$
So I think you can see where I'm going: the force between 2 charges:
$$ {\vec F} = \frac {q_1q_2} {4\pi\epsilon_0}\frac{\hat r}{r^2}$$
thus goes as the product of the charges.
A: This is a lot simpler than I think you are making it out to be.
Suppose the force on some $q$ due to some other $Q$ is $f.$ What would it be for twice the charge, $2Q$?
Well if superposition is to hold—in other words if the universe behaves in the simplest way—then the answer must be the same as “what is the force from two charges $Q$ each a hair’s breadth apart?” and the answer must be $f+f,$ or $2f.$ For $3Q$ we would get $f + f + f = 3f$. In general $nQ$ must lead to the force $nf$ for natural $n$.
One can immediately extend this to rational $r$ by reasoning backwards, or counterfactually: whatever the force from $Q/2$ is, call it $\varphi,$ and then realize that the earlier charge $Q$ is twice this and so by the foregoing argument that force is $f = 2\varphi$ hence $\varphi = f/2.$ From there one has reals as things which admit successive rational approximations, only the obvious extension of $rQ$ having force $rf$ works OK with being rationally approximated.
The $q Q$ approach is the only way to guarantee this multiplicative scaling and also Newton’s third law.
