Consider $$ u_t=u_{xx}+f(u)-w,~~w_t=\varepsilon (u-\gamma w).~~(*) $$
Consider the ansatz $$ (u(x,t),w(x,t))=(u(\xi),w(\xi)),~~\xi=x+ct, c\in\mathbb{R}. $$
Putting this ansatz into $(*)$, it is said that we get $$ u'=v\\ v'=cv-f(u)+w\\ w'=\frac{\varepsilon}{c}(u-\gamma w), $$ where $'=d/d\xi$.
I do not see that.
Consider $\frac{d}{dt}u(\xi)$, then the inner derivative is $c$, but what is the outer derivative?
The point of this is to convert the partial differential equations with functions of two variables into a system of ordinary differential equations with functions of one variable. It's a bit misleading to use $u(x, t) = u(\xi)$ because the $u$'s are actually different functions.
Suppose we define $u(x,t) = U(x + c t)$ and $w(x, y) = W(x + c t)$. Now the capital letters are functions of a single variable being applied to $\xi = x + c t$.
Looking at the first two equations, we have $u_{xx} = U''(x + c t)$, $u_t = c U'(x + c t)$, and $w_t = c W'(x + c t)$. Now you can rearrange the equations into:
\begin{align*} U'(\xi) &= U''(\xi) + f(U(\xi)) - W(\xi)\\ c W'(\xi) &= \varepsilon(U(\xi) - \gamma W(\xi)) \end{align*}
Using the substitution $V(\xi) = U'(\xi)$ you can get the system you stated. This is a first order nonlinear system or ODEs. Whether or not you can get a closed form solution will depend on the nature of $f$.