In Jackson's book on classical electrodynamics (3rd ed, ch 11, p. 516), he mentions how a wave equation for a field $\psi(\bf{x}^{'},t^{'})$ is transformed under Galilean shift, defined as $\mathbf{x}^{'}=\mathbf{x}-\mathbf{v}t$ and $t^{'}=t$, from a the frame $K'$ to the frame $K$ as
$$ \text{In frame $K'$: } \left( \sum\limits_{i}\frac{\partial^{2}}{\partial x^{'2}_{i}} - \frac{1}{c^{2}}\frac{\partial^{2}}{\partial t^{'2}}\right)\psi=0$$
$$ \text{In frame $K'$: } \left( \nabla^{2} - \frac{1}{c^{2}}\frac{\partial^{2}}{\partial t^{2}} - \frac{2}{c^{2}}\mathbf{v}\cdot\mathbf{\nabla}\frac{\partial}{\partial t} -\frac{1}{c^{2}}\mathbf{v}\cdot\mathbf{\nabla v}\cdot\mathbf{\nabla}\right)\psi=0$$
How was this latter equation derived?
The critical observation is that in the new coordinates, $\mathbf{x'}$ depends on $t$ and this needs to be taken into account in the $t$ derivatives. For example, to compute the first partial of $\psi$ with respect to $t$, we need to look at the following: $$ \begin{align} \frac{\partial}{\partial t} \psi(\mathbf{x} - \mathbf{v}t, t) &= \sum_i \frac{\partial \psi}{\partial x_i'} \frac{\partial x_i'}{\partial t} + \frac{\partial \psi}{\partial t'} \frac{\partial t'}{\partial t}\\ &= - \sum_i \frac{\partial \psi}{\partial x_i'} v_i + \frac{\partial \psi}{\partial t'} \\ \end{align} $$
I think that you can complete the calculation from there.