Numerical higher order derivatives and time axis
When considering finite differences there are three of them: Forward, backward and central.

Neither of them is "shorter" than the main signal. The zero is not exactly zero, it should be $\emptyset$. That is, "I don't know".

In backward difference, $y[n] = x[n] - x[n-1]$, but at $n=0$, $n-1$ is not even defined and the same applies for $y[0]$. We usually do $y[0]=0$, (which implies that $x[-1]=x[0]$).

Notice here that we wrote $y[n]$. In finite differences, in the discrete world, there is no $x[0.5]$. It is either $0$ or $1$. Central difference might appear to have a $\frac{1}{2}n$ term but that could be seen as a "phase shift" with respect to the sampling of the signal. The relative temporal difference between two successive samples is still one sampling period.

And this is the interesting bit here. The finite difference is the result of something that happens to two samples. Not one.

So, from the point of view of visualisation, you could show your first order finite difference as "occuring" between samples ($+\frac{1}{2}Ts$, where $Ts$ is the sampling period). The second order finite difference is something that "occurs" between pairs of samples of the first finite difference. The third order finite difference... (and so on).

Notice here that as you are working "up" the finite differences order, the starting point is "sliding" and the total length of the finite difference signal becomes shorter. The first order finite difference is between $x[n],x[n+1]$, the second order finite difference is between $x'[n],x'[n+1]$ which originate from $(x[n], x[n+1])$, $(x[n+1], x[n+2])$, the third order finite difference is between $x''[n], x''[n+1]$ which originate from $(x[0], x[1])$, $(x[1],x[2])$, $(x[3],x[4])$ and so on.

Notice here that throughout this nested evaluation of these finite differences that "I don't know" bit is becoming larger and larger. You cannot evaluate the $y'''[0]$ of the third finite difference if you don't know the values of the first four samples of $x[n]$. But if you consider an "overlapping" segment across a part of the signal that is long enough to evaluate all finite differences, then you would be able to "see" all of them.

Hope this helps.