Understanding the change of coordinates in differential operators

I'm reading the book "A Course on Partial Differentail Equations" by Walter Craig, and in the section 2.4 of the book the author claims the following:

"... sometimes called the Cauchy problem: $$ \partial_t^2 u - \partial_x ^2 u = 0, \\ u(0,x) = f(x), \quad \partial_t u(0,x) = g(x). $$ The initial data comprises two functions $f$ and $g$ that are to be specified. Introduce the new coordinates $$ x + t:=r, \quad x-t =s, $$ so that the differential operators are transformed as follows: $$ \partial_t - \partial_x = 2\partial_r, \quad \partial_t + \partial_x = -2\partial_s." $$ It's this change of coordinates that is not clear to me... I mean, why is that last sentence true? I'm guessing it's just something silly, but after thinking a bit, it was still not clear to me how to get this.

Not sure if that is a typo in the book or not, but here's what I remember from learning d'Alembert's formula. The solution in the original coordinate is $u(t, x)$. Introducing the new coordinates $r = x + t$ and $s = x - t$, we have $x = \dfrac{r + s}{2}$ and $t = \dfrac{r - s}{2}$ and so $$u(t, x) = u\left(\frac{r - s}{2}, \frac{r + s}{2}\right) = v(r, s).$$ It follows from Chain Rule that $$ \partial_tu = \partial_rv\cdot\frac{\partial r}{\partial t} + \partial_sv\cdot \frac{\partial s}{\partial t} = \partial_rv - \partial_sv. $$ Similarly, one can show that $\partial_xu = \partial_rv + \partial_sv$. In particular, \begin{align*} \partial_tu + \partial_xu & = 2\partial_rv \\ \partial_tu - \partial_xu & = -2\partial_sv. \end{align*} The wave equation transforms to $$(\partial_t - \partial_x)(\partial_t + \partial_x)u = -4\partial_{sr}v = 0. $$