Suppose we have $\xi=x-ct$ and $\tau=\omega t$ as well as $$ u(x)=u\left(\xi+\frac{c}{\omega}\tau\right)=:u(\xi,\tau). $$
What is then $$ \frac{\partial}{\partial\tau}u(\xi,\tau)? $$
I think this is $$ \frac{\partial}{\partial \tau}u(\xi,\tau)=(u_{\xi},u\tau)\cdot \begin{pmatrix}\frac{c}{\omega}\\0\end{pmatrix}=\frac{c}{\omega}u_{\xi} $$ by the multidimensional chain rule?
I find your notation somehow misleading, but the final result is, by the chain rule, $$ \frac{\partial}{\partial \tau} u(\xi,\tau)=u' \left(\xi + \frac{c}{\omega} \tau \right) \cdot \frac{c}{\omega}, $$ where $u'$ stands for the derivative of $u$.