Let $u(t,x)$ be a function in the coordinates $t$ and $x$. Now $u$ is to be expressed in coordinates $t$ and $\xi$ instead where $\xi=x-ct$.
Do we then have $$ u_{xx}=u_{\xi\xi}+cu_{\xi}? $$
I tried to answer this but did not come along. Maybe you can help me?
It helps to have different names for different things. If you write $$ U(t, \xi) = u(t, \xi + ct) $$ then you can compute \begin{align} U_{\xi}(t, \xi) &= u_x(t, \xi + ct) \cdot \frac{d(\xi + ct)}{d\xi} \\ &= u_x(t, \xi + ct) \end{align} and similarly, \begin{align} U_{\xi\xi}(t, \xi) &= u_{xx}(t, \xi + ct). \end{align}
Post-comment addition:
On the other hand, if you compute $U_t$, you get
\begin{align} U_{t}(t, \xi) &= u_t(t, \xi + ct) \cdot \frac{d(t)}{dt} + u_x(t, \xi + ct) \cdot \frac{d(\xi + ct)}{dt} \\ &= u_t(t, \xi + ct) + u_x(t, \xi + ct) c \\ \end{align} so that \begin{align} u_t(t, \xi + ct) &= U_{t}(t, \xi) - u_x(t, \xi + ct) c \end{align} And since $u_x = U_{\xi}$ (modulo shuffling arguments a little), this can be rewritten \begin{align} u_t(t, \xi + ct) &= U_{t}(t, \xi) - c ~U_{\xi}(t, \xi) \end{align}