How to go from $\frac{\partial^2 u}{\partial t^2}=a^2\frac{\partial^2 u}{\partial x^2}$ to a system of simpler equations?

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I am trying to understand my lecture notes.

We have $$\frac{\partial^2 u}{\partial t^2}=a^2\frac{\partial^2 u}{\partial x^2}.$$

For $a>0$, substitute: $\xi=x-at, \eta=x+at.$ Then

$u(x,t)=v(x-at,x+at)=v(\xi,\eta), \\\frac{\partial^2v}{\partial \xi \partial \eta}=0, \\ v(\xi,\eta)=\psi_1(\xi)+\psi_2(\eta).$

Then this notation is used:

$S(\xi)=\frac{\partial v}{\partial\xi}, R(\eta)=\frac{\partial v}{\partial \eta}.$ I don't understand why not write $S(\eta)=\frac{\partial v}{\partial\eta}$, but so it is in my notes. $s(t,x)=S(x-at), r(t,x)=R(x+at).$ I do not understand why we need this chain of different notations.

From this, a conclusion is made that the original equation is equivalent to this system:

$$ \begin{equation} \begin{cases} \frac{\partial r}{\partial t}+ a\frac{\partial r}{\partial x}=0\\ \frac{\partial s}{\partial t}-a\frac{\partial s}{\partial x}=0 \end{cases}\, \end{equation}. $$

I do not understand how we came to this conclusion. I see that $\frac{\partial^2 u}{\partial t^2}=a^2\frac{\partial^2 u}{\partial x^2}$ can be written as $\left(\frac{\partial }{\partial t }+ a\frac{\partial }{\partial x}\right) \left(\frac{\partial }{\partial t}-a \frac{\partial }{\partial x}\right)u=0$, but this has been of no help to me.

I would be very grateful if someone could give a detailed explanation. Thank you.