According to relativity theory, what is the force that two electrons moving radially apart exert on each other? According to relativity theory, what is the most general expression for the force that two electrons moving radially apart exert on each other?
I am looking for a function $$F(r(t)),$$ where $F$ is the mutual force and $r(t)$ is the electrons' separation at time $t$.
N.B.: The bounty will go to the derivation of the most general possible expression, not just the case when $\ddot{r}(t)=0$.
 A: The force between the charges goes to zero. 
To see this, work in the frame of one of the charges. From its perspective, the other point charge is moving rapidly away, and the field of a moving charge is weaker along the direction of motion, as shown below.

One cheap way of seeing this is to pretend the field lines have been "length contracted". For arbitrarily high velocities, the field parallel to the velocity gets arbitrarily small, so the force vanishes.

We can also be a little more quantitative. In the lab frame, let the particles have velocity $v$ and Lorentz factor $\gamma$. The field one particle exerts on the other is
$$\mathbf{E} \propto \frac{\gamma}{r^2 (1+\gamma^2 v^2/c^2)^{3/2}} \hat{r}.$$
Taking the $v \to c$ limit, $\gamma \gg 1$ and we have
$$\mathbf{E} \propto \frac{\gamma}{r^2 (\gamma^2)^{3/2}} \hat{r} \propto \frac{1}{\gamma^2} = 1-v^2/c^2 \propto 1-v/c.$$
where we used the limits several times. So if you start with huge velocities and get them 50% closer to the speed of light, the force is cut in half.
A: The solution to this interesting question has to involve both (a) the distortion of the electric field of point charges when they move close to the speed of light and (b) time (since the longer we wait the further apart the electrons become, so their mutual force becomes smaller).
Since the electrons are moving along the same straight line we can reduce this problem to one of finding the electric field along a single axis (the x-axis).
Also since your question asks only if the force tends toward zero we don't need to actually include the motion of both electrons - we can calculate the force at the central position due to one of the electrons and if that tends toward zero then the force on the other electron will also tend toward zero (since it is further away than the central position).
According to equation 11.152 of Jackson's book Classical electrodynamics, the electric field at x=0 due to a particle with charge $q$ moving in the x-direction away from x=0 with velocity $v_x$ is:
$E_x=-\frac{q}{\gamma^2 (v_xt)^2}$
As the particle approaches the speed of light, $\gamma$ approaches infinity and so the electric field approaches zero. The other electron is moving away from the origin so the field at its location will also approach zero.
Obviously, as time increases there is a further factor that reduces the electric field at the origin (sine the electron is moving away from the origin) and this should be true even in the non-relativistic case $\gamma \approx1$.
Second edit to account for new question from the OP:
To get the field at the location of the other electron is a little more involved. We use "primes" to indicate variables in a coordinate system moving with the charge that moves to the right, and unprimed coordinates refer to the lab frame (in which electrons move away at equal speed in either direction). In the frame moving with the electron that moves to the right, the electric field can be found from the gradient of the potential, which is:
$\phi'=q/r'$
and since we already mentioned we only need the x-axis, we can take $r'=x'$. So $\phi'=q/x'$. The electric field is then (in the primed frame):
$E_x'=-\partial/\partial x'(\phi') =-\frac{q x'}{(x'^2)^{3/2}}$
using the Lorentz transform $x'=\gamma (x-vct/c)$ in the above gives:
$E_x'=\frac{q \gamma (x-vct/c)}{(\gamma (x-vct/c))^{3}}$
Since the electric field in the direction of motion is not transformed, $E_x'=E_x$, so
$E_x=\frac{q \gamma (x-vct/c)}{(\gamma (x-vct/c))^{3}}$
note that this expression reduces to the one above when we take $x=0$ as we did above. However, we want the field at the location of the other electron, which, in the lab frame, has position $x=-vt$. Inserting this in the above expression for the field gives:
$E_x=\frac{q }{4\gamma^2 (vt)^2}$
The factor 4 comes from $(2vt)^2$ and reflects the fact that the relative velocity of the two electrons is $2v$.
Third edit to account for new question from the OP:
If the charges are accelerating, then since the force is directed along one axis only, we don't need to include the acceleration term in the field expression (see the third term in equation 28.3 of Feynman Vol II).
The velocity of the instantaneous rest frame of the particle moving to the right is $v=v_0+at$, where $v_0$ is the velocity at $t=0$ and $a$ is its acceleration.
We take the result derived above:
$E_x=\frac{q }{(\gamma(v) (x-vt))^{2}}$
and note that $\gamma=\gamma (v)$. Now, the particle moving to the left has location $x=-at^2/2-v_0 t$. Inserting these new expressions for $v$ and $x$ into the above equation gives:
$E_x=\frac{q }{(\gamma(v) ((-at^2/2-v_0 t)-(v_0+at)t))^{2}}$
which reduces to the non-accelerating case when $a=0$.
