The pulses of light could be considered a tick-tock of a clock, right?
Kind of... if you account for the effect of the finite speed of light propagation. Namely, the rate at which the Receiver sees the pulses arrive is not the rate of the Sender's clock in Receiver's reference frame.
If so, then what the moving Receiver is observing is that the Sender's clock has slowed down.
That's what Receiver observes - i.e. can carefully conclude from observation. But what he sees is pulses arriving at an accelerated rate, giving a first (incorrect) impression of a faster-ticking Sender's clock.
Now, here's where the mind-**** happens for me. While stationary relative to Sender, Receiver puts a rod behind him that is one light-second long. That means they both agree it is 186,000 miles long from A (Receiver's ass) to B (end of the rod). Correct?
So, while still stationary relative to one another, both will agree that the laser will hit points A and B one second apart. Correct? If we imagine the laser beam looking like a Star Trek phaser, they will be the same length.
With you so far.
When Receiver is moving at 0.5c relative to Sender, then Sender will observe it hitting points A and B 0.87 seconds apart. Thus, the length of the rod has shortened due to length contraction. Sender still measures light moving at c.
Actually, no. The rod is indeed shortened in Sender's frame and measures 0.866 light seconds, but Sender will not observe the light hitting points A and B 0.866 seconds apart, because the rod is moving.
Sender will observe the light hitting points A and B 0.577 seconds apart.
Now, if there was a mirror at point B, and the light was reflected, Sender would see that it takes light 1.732 seconds to get from B to A, because the 0.866 light seconds long rod is "flying away" from the returning beam of light.
Altogether, Sender will see that it took 2.309 seconds for light to get from A to B to A. (The asymmetry of the legs' durations in Sender's frame is due to tilting of the plane of simultaneity over the length of the rod.)
What about the Receiver? The rod is stationary relative to Receiver, so he still measures it as 186,000 miles long. It will still take one second for it to go from point A to point B.
Yes. And one second to go back again, if there's a mirror at point B.
Thus, if Receiver times how long it takes the light to return to him, he measures 2 seconds, while Sender observes that it took 2.309 seconds (which happens to be 2/0.866) - and that Receiver's clock has slowed down, advancing only by 2 seconds during those 2.309 seconds.
(Just for completeness: Receiver's take would be that Sender, when measuring those 2.309 seconds, did not start and stop his stop-clock at the correct moments that light passed through A. If Sender had started and stopped his stop-clock at the moments that Receiver would consider simultaneous with light passing through A, Sender's clock would have measured 1.732 seconds. Receiver would therefore observe that Sender's clock has slowed down, advancing only by 1.732 seconds during a 2 second interval.)
Only now when it reaches point B, the "tail end" of the laser beam will have passed point A 0.13 seconds ago as observed by Receiver.
I'm not completely sure what you mean here. In general, relativity causes different-moving observers to disagree about the simultaneity of events that are not colocal (that occur at different places).