Realising I don't truly understand differentiation when not stated explicitly.

I often find it to be more clear to write inputs explicitly, and also to use the notation $D_i f$ for the $i$th partial derivative of a function $f$. The function $u$ is defined so that $$ u(x,t) = v(x + ct, x - ct) = v(\xi(x,t), \eta(x,t)), $$ where $\xi(x,t) = x + ct$ and $\eta(x,t) = x - ct$. From the multivariable chain rule, the first partial derivative of $u$ is \begin{align} D_1 u(x,t) &= D_1 v(\xi(x,t), \eta(x,t)) D_1 \xi(x,t) + D_2 v(\xi(x,t), \eta(x,t)) D_1 \eta(x,t)\\ &= D_1 v(x + ct, x - ct) \cdot 1 + D_2 v(x + ct, x - ct) \cdot 1. \end{align} This result could also be written as $$ \partial_x u(x,t) = \partial_\xi v(x+ct, x - ct) + \partial_\eta v(x + ct, x- ct). $$

---

Regarding the second example, define $$ F(x,t) = f(ax + bt) = f(h(x,t)), $$ where $h(x,t) = ax + bt$. From the chain rule, we have \begin{align} D_1 F(x,t) &= f'(h(x,t)) D_1 h(x,t) \\ &= f'(ax + bt) \cdot a. \end{align} This calculation could also be written as \begin{align} \partial_x F(x,t) &= f'(h(x,t)) \partial_x h(x,t) \\ &= f'(ax + bt) \cdot a. \end{align}