The textbooks says that $$ \frac{\partial u}{\partial t} + c \frac{\partial u}{\partial x} + u \frac{\partial u}{\partial x} = 0$$ can be transformed into $$\frac{\partial u}{\partial t} + u\frac{\partial u}{ \partial x'} = 0$$ by introducing a new variable $x' = x - c t$.
It's embarrassing, but I can't do it. It seemed logical to me at first glance to employ a chain rule here like $\frac{\partial u}{\partial x} = \frac{\partial u}{\partial x'} \frac{\partial x'}{\partial x} = \frac{\partial u}{\partial x'}$, but obviously something else has to be done.
Understanding the two PDEs
In the first equation you have $u(x,t)$ so $\partial u/\partial t$ means differentiate $t$ and hold $x$ constant. In the second equation we hold $x'$ constant, when differentiating with respect to $t$.
Making the change of variables
I will now use a notation that my undergraduate physics professor Dr. Buchholtz used in his statistical mechanics course, in which a subscript is used to denote which variable is being held constant when differentiating.
\begin{align*} \left(\frac{\partial u}{\partial t}\right)_{x}&=\left(\frac{\partial u}{\partial t}\right)_{x'}+\frac{\partial u}{\partial x'}\left(\frac{\partial x'}{\partial t}\right)_{x}\\ &=\left(\frac{\partial u}{\partial t}\right)_{x'}-c\frac{\partial u}{\partial x'}\\ \end{align*}
Using the following that you were able to determine
\begin{align*} \left(\frac{\partial u}{\partial x}\right)_{t}&=\left(\frac{\partial x}{\partial x'}\right)_{t}\\ \end{align*}
will allow you to properly make the change of variables and obtain the result from the book.
I hope this helps.