How are the time dilation and the relativistic Doppler shift added together in what observers "see"? I'm using the illustration from this question:

Suppose A and B are d light years away, and at rest. Then they symmetrically start to travel toward each other (symmetric acceleration process in negligible time) at a high speed v. How each one sees the other one's clock working? 
"Always faster" and "always slower" are both logically impossible, since due to the symmetry, they must agree that same time has elapsed on both clocks when they meet. So it's either "always normal" or probably "faster for some time and slower for some time". What's the mathematical expression for the seeming rate of clocks?
 A: There is some difference between “seeing” and “calculation” or “interpretation”.
At relativistic velocities you must include time dilation into Doppler Shift, either to A-clock or to B-clock or, as in your case - to the both. That does not affect measured frequency shift, but changes judgment regarding the speed of ticking of the "other clock".
Approaching each other the both will "see" or measure very high frequency, so they will "see" each other clock  ticking very fast. However, that says nothing about “actual” clock rate of any of them; it is frame dependent and is a matter of interpretation.
Relativistic Doppler Effect is very simple – it is “ordinary” Doppler shift plus time dilation. If you think that B is moving in the frame of A, then you attach time dilation to B. If you think that A is moving in the frame of B, then you attach time dilation to A. If you think that they both move with equal but opposite velocities, then you can share full amount of time dilation – Lorentz factor in equal proportions. Measured blueshift of frequency will be the same, but contributions of time dilation will be frame dependent.
Let's B is emitting monochromatic light towards A. For example, one may think like that: A is at rest and B is moving towards A with velocity very, very close to c. B catches up all emitted by him wave fronts and emits new ones; hence the wavefronts gather straight in the front of him. These wavefronts will hit A almost at once, like fighter jet sonic boom. A will see very intense blueshift of frequency, but due to dilation of B clock it will be less intense than it could be in the classical case.
Or you may think that B is at rest and A is moving towards B. In this case maximum observed by A frequency (in classical case) should be $2f$  (wavefront is moving with velocity c and A is moving towards wavefronts with velocity close to c). But, due to dilation if A’s clock A will see this frequency as very intense (blueshift). A will explain it this way: my clock is ticking very slowly; I have turned into dawdler; that’s why I see all processes around me as very, very fast.
In your exact case every observer will explain or interpret that the other clock is ticking at the same rate as his own; B clock dilates and A clock dilates at the same magnitude and that time dilation cancel each other; so the measured Doppler Shift is no different from classic one.
So, since in your case they move with the same but opposite velocities, their clocks slow down at the same rate indeed.
Please look for relativistic Doppler Shift in Feynman Lectures (Relativistic Effects in Radiation).
Finally, if the both move at parallel lines, at certain moment (points of closest approach) they may see purely clock rate of each other clock – Transverse Doppler Effect. For "moving" source - light emitted at points of closest approach redshifs ; for "moving observer" - light received at points of closest approach blueshifts.
Sure, they will measure zero - Transverse Doppler Effect for the frame, in which they move with equal but opposite velocities,
A: One important assumption that I've omitted is that the travelers have had been at rest for long enough. Additionally let's assume they have synchronized their clocks. 
"Always faster" is logically possible, and is indeed the case. How? Suppose $d=10$ light years, and $v$ is so that the travelers meet after 2 years, according to their own clocks. At the starting point of the journey, each one observes the other one's clock showing $R1=-10$ years. When they meet, the second reading will be $R2=2$ years. Therefore, the average observed rate of clocks will be $12/2=6x$. 
More specifically, each one observes a step function: A constant rate of $S1$ for $t$ years, which jumps to $S2$, and remains constant for the remaining $2-t$ years. I'm not skilled enough to calculate $t$, $S1$ and $S2$, but obviously, $S2>S1>1$ and $(t*S1+(2-t)*S2)/2=6$. 
