We have the following equality:
$$ u(x,t) = v (\xi, t)$$ with $\xi = x - ct$
In my textbook it says that:
$$ u_t = \dfrac{ \partial v}{\partial \xi} \cdot \dfrac{ \partial \xi}{\partial t} + \dfrac{ \partial v}{\partial t} \cdot \dfrac{ \partial t}{\partial t}$$
I have never encountered this rule and I'm can't derive it. Is anyone able to clear this up?
Assuming you mean $$u(x,t)=v(\xi(x,t),t)$$
you would want to use the chain rule. It says indeed that
$$D_2 u(x,t)=D_1v(\xi(x,t),t)\cdot D_2\xi(x,t)+D_2v(\xi(x,t),t)\cdot 1$$ since $D_2(t)=1$, where $D_i$ means differentiation w.r.t. the $i$-th coordinate. In this case, your coordinates are $(x,t)$.
In general, suppose $f:\Bbb R^m\to\Bbb R^n$, and each $g_i:\Bbb R^n\to\Bbb R$.
Set $F:\Bbb R^n\to\Bbb R$ where $$F(x_1,\ldots,x_n)=f\circ g(x_1,\ldots,x_n)$$ and the function $$g=(g_1(x_1,\ldots,x_n),\ldots,g_m(x_1,\ldots,x_n))$$ is differentiable (or assuming each coordinate is continuously differentiable)
Then $$D_iF(x_1,\ldots,x_n)=\sum_{j=1}^m D_jf(g(x_1,\ldots,x_n))\cdot D_ig_j(x_1,\ldots,x_n)$$
