If someone left at $t$ and you leave H hours later, why your time is $t - H$? Not $t + H$? Why $\color{lightgreen}{-2}$ in this answer? Why not $40t = 60(t \color{red}{+ 2})$? Why doesn't later mean +?

The time when the two trains will meet is going to be the solution to the following equation (the intersection of two straight lines) where $t\ge0$ and $t=0$ corresponds to $12:00$ PM (noon):
$
40t=60(t\color{lightgreen}{-2})\implies\\
40t=60t-120\implies\\
t=6\ P.M.
$

 A: I feel that this question is best answered by an example. Picture yourself starting a timer as soon as you wake up in the morning — say you start the timer at 08:00. Let $t$ represent the number of minutes that the timer has been running throughout the day. Once you finish with breakfast, imagine that you decide to go for a walk. Before leaving, you check the timer and see that $t_1$ minutes have elapsed. Upon arriving back home, you check the timer, and you find that $t_2$ minutes have now passed. How can you determine the length of time you walked? Of course, you would take $t_2 - t_1.$
Letting $t$ vary now, if you want to know the duration of time that has passed since observing the timer at some number $t_1$ of minutes, you would use the function $f(t) = t - t_1.$ Observe that this is consistent with (1.) the fact that no time has elapsed the moment you look at the clock (i.e., $f(t_1) = 0$), and (2.) the duration of your morning walk is $f(t_2) = t_2 - t_1.$ Further, if $t_1 = 0,$ then $f(t) = t,$ so this is consistent with the fact that the amount of time that has elapsed since beginning the timer at 08:00 is $t$ minutes, as we defined it above.
A: It is so that both trains start at "zero" hours of travel. The first train starts actually at 0 and the second train when t=2 hours. In order to make its distance make sense, the speed (60) must be multiplied by the time it has been traveling. If it leaves at t=2 hours, then you must subtract those two hours so the the distance is correct
