I grasped my head around this topic a lot and I want to ask if my interpretation of the problem is now correct in the most clear way.
I'm trying to measure time dilation using the spacetime interval but adding the ingredient of the proper time, the measurable time, being compared between two observers (this part is usually overlooked).
Suppose two observers 1 and 2 in Minkowski spacetime ($c=1$). From the perspective of 1, 2 moves in a given direction with a velocity v. Assume 1 and 2 are colinear with the velocity vector. Two events happen: 2 passes through a point A and then through a point B. The interval of the events in the reference system of 2 is
$\Delta S^2$=-$\Delta t^2$
As 2 is at rest for the events ($\Delta x=0$), we say that the time measured by him is the proper time as he is the proper system of these two events ($\Delta \tau$=$\Delta t$). So in this case,
$\Delta S^2$=-$\Delta \tau^2$
Now, from the perspective of 1, 2 is moving with a velocity v in x'-direction. The interval of the events in the reference system of 1 (coordinates are x' and t') is
$\Delta S^2=-\Delta t'^2+\Delta x'^2 = -\Delta t'^2+v^2 \Delta t'^2= -(1-v^2)\Delta t'^2$
Common approaches stop here and say that, since $\Delta S^2$ is an invariant then
–$\Delta \tau^2=-(1-v^2)\Delta t'^2$
$\Delta t'=\gamma(v) \Delta \tau$
so there we have it. I believe we are missing something because $t'$ is a coordinate and not measurable time. Now I want to do a seemingly obvious step further that I think clarifies this result. I believe that, since we already used a proper time for this event, we have to take into consideration another event that happens and is related to this one that can lead us to use the time measured by 1. The perfect example is the one thing why I believe some people have a hard time understanding measurement in relativity: We are not taking into explicit consideration what is measuring time by an observer outside the proper system. That example is a series of events: the light reaching the observer's light cone. Since 1 and 2 are colinear with the movement of 2 (more simple), then we can talk about the second pair of events: 1 receives a light ray that is sent from 2 when he passes through A and another one that is sent when 2 passes through B.
Now the special part: Since 1 is at rest with these two events, he is now the proper system and measures proper time $\Delta \tau'$. The first light ray travels a certain distance and the second one travels an extra $c\Delta t'$ ($\Delta t'$ since $c=1$). So the difference between the light rays (the interval between the events) measured by 1 is $\Delta t'$, which corresponds to the total time measured $\Delta \tau'=\Delta t'$. Now we have the result
$\Delta \tau'=\gamma(v) \Delta \tau$
We now compare two measurable times and now we can say that 1 measure that 2 takes a longer time reaching B from A that the time it takes for 2 to pass through those points. It is the same result, but I believe this completes the picture.
Best Answer
$t'$ corresponds to the time on observer $1$'s wristwatch, and is therefore perfectly measurable. More precisely, in the reference frame of observer $1$, the two events "my stopwatch reads $t'$" and "observer 2 passes through point $B$" are simultaneous.
The reason we don't take that into account is because the actual mechanism by which observer $1$ concludes that the two aforemented events are simultaneous is irrelevant. There is no speed-of-light delay which needs to be accounted for here. To put things in the rod-and-clock language, we imagine that observer $1$ has placed synchronized clocks at every point in space, and $t'$ is the time on the clock at point $B$ when observer $2$ passes it.
I don't exactly follow what you're doing. If observer 2 sends light signals to observer 1 when they pass points A and B respectively, then the time between observer 1 receiving those two light signals is equal to $\gamma(v) \Delta \tau(1+ v/c) = \Delta \tau \sqrt{\frac{1+v/c}{1-v/c}}$. This calculation arises e.g. when computing the relativistic Doppler effect.