I see this thread has already gotten very long, and I haven't had time to read it, alas. But there is a fundamentally simple point that I'd like to make, though others may have made it before.
That point is that the process of comparing clocks in an inertial frame is frame dependent.
It is convenient to compare clocks using the Einstein synchronization convention, as it is very standard, and most schemes are equivalent. Operationally, one way of describing this is that a light signal is emitted from the midpoint in the inertial frame, then the light signal is received "at the same time" , when it reaches its destination.
As is explained by the "Einstein's Train" thought experiment, this process is frame dependent. Google for it, or for "the relativity of simultaneity". This is one of the trickiest parts of SR to get across, by the way.
This is a necessary insight to understand how A can think B's clock is slow, and B can think A's clock is slow.
Since it was mentioned that the OP is not afraid of math, I'll go through an abstract argument about why symmetrical time dilation implies simultaneity must be relative. It's not much math - it's just the abstract notion of an invertible map, a 1:1 correspondence between sets, also known as a bijection. It may be overkill, but there's a wiki article on this at
https://en.wikipedia.org/wiki/Bijection. For every element in one set (say A) there is one unique corresponding element in set B. And since the mapping is invertible, for every element in set B, there is one, and only one, corresponding element in set A.
The necessary assumption is that comparing times is done by such a one-one map, by a bijection.
Let's take a specific example. Let's say, for simplicity, that A thinks B's clock is running at half seed, and B thinks A's clock is running at half speed.
If there is only one mapping from A's time to B's time, this is logically impossible. With only one mapping, if B's clock is running at half A's rate, then A's clock is running twice as fast as B's clock.
A=1 corresponds to B=2. Logically, B=2 must correspond to A=1, as the map is unique and invertible - it's unique in both directions.
However, if the mapping from A's time to B's time is different from the mapping from B's time to A's time, this is perfectly possible.
In special relativity, this is the case. The mappings are done by clock syncrhonizattion convetions. A is in A's frame, and A uses the convention for this frame to make A's map. B is in B's frame, and B uses frame B's convention to make B's map. The important thing to realize is that A's map is NOT THE SAME as B's map.
If necessary, we can talk more about why we use Einstein's synchronization convention, but at this point it would distract from the main point, I think.