Why does it take $2d/\lvert v_1 - v_2\rvert$ time to overtake? (Can physics catch-up to math?) The following answer is from this post.

So it becomes particularly easy to see that the opposite-direction meetings must occur at the interval $2d/(v_1 + v_2)$ and that the overtaking meetings must occur at the interval $2d/\lvert v_1 - v_2\rvert,$
since those are the relative slopes of the red line and the grid lines in each case and the grid lines are spaced vertically $2$ road-lengths apart.

Which was addressing the following question from the same post:

$2$ cars travelling at $108,75$ kmph respectively, start from the opposite ends of a $90$ km straight road. Travelling back and forth continuously for $10$ hours, how many times do the $2$ cars meet? [No loss of time at the turns]

Can somebody please explain the physics behind the mathematically attained $2d/\lvert v_1 - v_2\rvert$?
Firstly, while overtaking I don't think that the distance $2d$ should appear in the numerator (in that expression). Also, even if we do consider $2d$ still not all of that would be covered at the relative speed of $\lvert v_1 - v_2\rvert$ (as some of it is on opposite direction too).
But apparently that's not the case, it is working totally fine. Since the answer works so I'm confused how, instead of math, physics explains the above?
 A: I think the the answer you linked, written by David K, does give a lot of physical insight, by essentially applying the method of images. The formula in the answer you cited gives the times between successive same-direction meetings. To understand where that comes from, it might be interesting to think through how to derive the times of all same-direction meetings, starting from David K's application of the method of images.
Let's look at the last figure produced in the answer by David K, which I think is quite helpful.

The "same direction" intersections will occur when the red line crosses a positive sloping black line.
We will assume $v_1 > v_2$, so in this figure $v_1$ is the slope of the red line and $v_2$ is the slope of the black line.
The red line has the formula
\begin{equation}
x_1(t) = v_1 t
\end{equation}
The first positive sloping black line has the equation
\begin{equation}
x_{2, 1}(t) = d + v_2 t
\end{equation}
Setting $x_1(t) = x_{2, 1}(t)$ equal allows us to solve for $t_1$, the time of the first same-direction intersection
\begin{equation}
t_1 = \frac{d}{v_1 - v_2}
\end{equation}
As a numerical sanity check, note that in the units of the figure, $d=1$, $v_1=1.2$, and $v_2=0.83$, therefore $d/(v_1-v_2) \approx 2.7$, which seems to agree with where the red line intersects the first upward sloping black line. This value also aligns with the value $2.72$ hours in @James's table.
The next positive sloping black line has the equation
\begin{equation}
x_{2, 2}(t) = 3 d + v_2 t
\end{equation}
The reason that the second line starts at $3d$, instead of at $2d$, is due to the reflecting boundary condition. Cars separated by $2d$ are actually in the same location.
Solving for $t_2$, the time of the second same-direction intersection, yields
\begin{equation}
t_2 = \frac{3d}{v_1 - v_2}
\end{equation}
As a sanity check, using our values from before, we'd expect $t_2 \approx 3 / (1.2-0.83) \approx 8.1$, which again looks consistent with the graph. This value also aligns with the value $8.17$ hours in @James's table, given the level of precision to which we are working. (And adding more decimal places to $0.83333...$ makes the agreement better).
In general, the position of the $k$-th black line (where $k=1, 2, 3, \cdots$) is
\begin{equation}
x_{2, k} = (2k-1) d + v_2 t
\end{equation}
so the time of the $k$-th same-direction intersection $t_k$ will occur at time
\begin{equation}
t_k = \frac{(2k-1)d}{v_1 - v_2}
\end{equation}
Since $t_3 \approx 13.6 > 12$, there are only two same direction crossings in the first $12$ hours.
Note that, consistent with the answer you linked to,
\begin{equation}
t_k - t_{k-1} = \frac{2d}{v_1-v_2}
\end{equation}
One way to understand this is that due to the reflecting boundary condition, we need to space the "copies" of the original cars by a distance $2d$ along the y axis in the figure above.
