Why is rapidity additive?

Relativistic velocity addition takes the form $$ V = \frac{v+u}{1+uv/c^2} $$ The rapidity is defined so that $$ v = c \tanh \eta $$ Plugging this into the the velocity addition formula and noting that $V = c \tanh N$, we find $$ c \tanh N = c \frac{ \tanh \eta + \tanh \eta' }{1 + \tanh \eta \tanh \eta'} = c\tanh(\eta+\eta') $$ Thus, the new rapidity is $$ N = \eta + \eta' $$

Alright here's a very crude heuristic(definitely not a proof) on why(without using velocity composition) rapidity is additive.

Impose these two constraints:

1) the constancy of speed of light in all frames. This means that the rapidity of light $\tanh\phi_c=1$ is the same in all frames of reference.

Consider this case:

You observe a frame of reference $S'$ moving with $\tanh \phi_1=v$ with respect to $S$ followed by a light beam with $\tanh\phi_c=1$. we know the rapidity of light in the $S'$ frame is the same as in $S$. Given this fact we'd like to find out what is the relation $ f(\phi_1,\phi_c)$ between the old and new rapidities in the new frame $S'$ such that: $$\tanh( f(\phi_1,\phi_c))=1$$

2) The second constraint is that the rapidity of an object at rest is zero. So if there's an object moving with velocity $v$ in $S$, it's at rest in $S'$. That is $$\tanh( f(\phi_1,\phi_1))=0$$

Then it's reasonable to conjecture that $f(\phi_1,\phi_2)=\phi_2 -\phi_1$

As you can check yourself using the hyperbolic identity $$\tanh (\phi_1 +\phi_2)=\dfrac{\tanh \phi_1+ \tanh \phi_2}{1+\tanh \phi_1 \tanh \phi_2} $$

Update: I was asked in the comments by user Prahar that I jumped into the conclusion too quickly. why for instance the functions $$(\phi_2 -\phi_1)^2$$ Or say $$(\phi_2 -\phi_1)^n$$ Or say $$\ln(\dfrac{\phi_2}{\phi_1})$$ are invalid?Although they satisfy the first two constraints?

The reason has to do with another constraint:

3)The principle of relativity dictates the following: If you're $S$ and $S'$ is moving with $\tanh \phi_1 =v$ according to $S$, then according to $S'$, $S$ is moving with $-v$ so that $$\tanh (f(\phi_1,0))=-v$$ This is only satisfied for $$\phi_2 -\phi_1$$

Because for $$(\phi_2 -\phi_1)^2$$ or any other $n$, We have $$\tanh ( (f(\phi_1,0))= \tanh (0 -\phi_1)^2= \tanh \phi_1^2 $$ which is evidently not equal to $-v$ because $\tanh \phi_1^2 $ is positive(whereas it should have been negative) and also because it's never equal to $v$, since $\tanh$ is a one to one function.

While looking into it again I've also found this simple solution to the problem.

Rewriting the Lorentz-Transformations with the rapidities gets us this: $$x'=x\cosh{\phi}-ct\sinh{\phi}$$ $$y'=y$$ $$z'=z$$ $$ct'=-x\sinh{\phi}+ct\cosh{\phi}$$

From this, it's easy to find:

$$ct'\pm x'=e^{\mp\phi}(ct\pm x)$$

Simply writing this out for $S\rightarrow S'$ and $S'\rightarrow S''$ and then substituting will get you:

$$ct''\pm x''=e^{\mp(\phi_2+\phi_1)}(ct\pm x)$$