If I change coordinates from $(x,t)$ to $(u,v)$ via the formulas $u=x-ct$ and $v=x+ct$ then I want to show that $$\frac{\partial}{\partial x} - \frac{1}{c}\frac{\partial}{\partial t}=2\frac{\partial}{\partial u}$$ $$\frac{\partial}{\partial x}+\frac{1}{c}\frac{\partial}{\partial x}=2\frac{\partial}{\partial v}$$
but I'm not sure how to do this.
On
Somebody can offer a more comprehensive explanation, but I try to do. The operators, like these, are defined by its action over some suitable set of objects. Thus its properties (or relations) have to be proved taking instances of those objects and making the operators act on them. The partial derivative acts over functions following the known rules for derivation. Equalities for partial derivatives are true if are true these equalities when acting over any function.
$$\frac{\partial \psi (x(u,v),y(u,v))}{\partial x}=\frac{\partial \psi}{\partial u}\frac{\partial u}{\partial x} + \frac{\partial \psi}{\partial v}\frac{\partial v}{\partial x}=\frac{\partial \psi}{\partial u}\ + \frac{\partial \psi}{\partial v}$$ $$\frac{\partial \psi (x(u,v),y(u,v))}{\partial t}=\frac{\partial \psi}{\partial u}\frac{\partial u}{\partial t} + \frac{\partial \psi}{\partial v}\frac{\partial v}{\partial t}=-c\frac{\partial \psi}{\partial u}\ + c\frac{\partial \psi}{\partial v}$$
$$\frac{1}{c}\frac{\partial \psi}{\partial t}=-\frac{\partial \psi}{\partial u}\ + \frac{\partial \psi}{\partial v}$$
$$\frac{\partial \psi}{\partial x}+\frac{1}{c}\frac{\partial \psi}{\partial t}=2\frac{\partial \psi}{\partial v}$$
$$\frac{\partial \psi}{\partial x}-\frac{1}{c}\frac{\partial \psi}{\partial t}=2\frac{\partial \psi}{\partial u}$$
From $u = x-ct$ and $v = x+ct$ we get $$ x = \frac 12 (u+v), \quad t = \frac 1{2c} (v-u). $$ Hence, by the chain rule $$ \def\pd#1#2{\frac{\partial #1}{\partial #2}}\def\p#1{\pd{}#1}\p u = \pd xu \p x + \pd tu \p t = \frac 12 \p x - \frac 1{2c} \p t $$ and $$ \p v = \pd xv \p x + \pd tv \p t = \frac 12 \p x + \frac 1{2c} \p t. $$