Let's first look at a scalar field since it has the most simple transformation properties. As the most simple example consider a free Klein-Gordon field fulfilling the Klein-Gordon equation
$$(\Box+m^2)\phi(x)=0,$$
where
$$\Box=\partial_{\mu} \partial^{\mu} = \partial_t^2-\vec{\nabla}^2.$$
I use the west-coast convention of the metric ("mostly minus") and set ##c=1##.
The transformation property of the scalar field under a Lorentz transformation ##x'=\hat{\Lambda}x## is
$$\phi'(x')=\phi(x)=\phi(\Lambda^{-1} x').$$
Now consider a plane wave solution, i.e., we make the ansatz
$$\phi(x)=\phi_0 \exp(-\mathrm{i} k \cdot x).$$
Plugging this into the Klein-Gordon equation gives the dispersion relation
$$k^2-m^2=0 \; \Rightarrow \; k^0=\omega=\sqrt{\vec{k}^2+m^2}.$$
The ##k## is a Minkowski four-vector (since the dispersion relation is a covariant equation!). Thus you have
$$\phi'(x')=\phi(\hat{\Lambda}^{-1} x')=\phi_0 \exp(-\mathrm{i} k \cdot \hat{\Lambda}^{-1} x').$$
Now Minkowski products are scalar, i.e., you can write
$$k \cdot \hat{\Lambda}^{-1} x' = (\hat{\Lambda} k) \cdot (\hat{\Lambda} \hat{\Lambda}^{-1} x')=(\hat{\Lambda} k') \cdot x',$$
i.e., you get again a plain wave in the new frame with the wave four-vector
$$k'=\hat{\Lambda} k$$
Of course it also fulfills the dispersion relation, as it must be:
$$(k')^2=(\hat{\Lambda} k)^2=k^2=m^2.$$
Obviously the amplitude is a scalar, i.e., it doesn't change:
$$\phi_0'=\phi_0.$$
Now take a boost in ##x## direction
$$t'=\gamma (t-v x), \quad x'=\gamma(-v t+x), \quad y'=y, \quad z'=z.$$
As we've seen the wave vector transforms in the same way, i.e.,
$$k^{\prime 0}=\omega'=\gamma(\omega-v k_x), \quad k_x'=\gamma(-v \omega+x), \quad k_y'=k_y, \quad k_z'=k_z.$$
This formulae include the Doppler effect, i.e., the frequency changed as measured in the new frame compared to the old frame and the socalled "aberration", given by the change of the spatial components.
For the electromagnetic field it's quite similar. You have only to consider that it is a vector field, given by the 2nd-rank Faraday tensor, i.e., the transformation properties are
$$F^{\prime \mu \nu}(x')={\Lambda^{\mu}}_{\rho} {\Lambda^{\nu}}_{\sigma} F^{\rho \sigma}(x)={\Lambda^{\mu}}_{\rho} {\Lambda^{\nu}}_{\sigma} F^{\rho \sigma}(\hat{\Lambda}^{-1}x').$$
If you work this transformation out in terms of the usual representation of the field in terms of electric and magnetic components ##\vec{E}## and ##\vec{B}## you get the transformation explained in
https://en.wikipedia.org/wiki/Class...rmation_of_the_fields_between_inertial_frames
If you put in plane waves you get of course massless dispersion relations, ##k^2=0 \; \Rightarrow \; \omega^2=\vec{k}^2##, and the Doppler shift and aberration formulae are the same as for scalar fields since ##k## is of course again a four vector. The polarization properties, of course also change according to the formulae given in the Wikipedia article.
The em. field has two invariants (scalar and pseudoscalar). In conventional formulation they are given by ##\vec{E}^2-\vec{B}^2## and ##\vec{E} \cdot \vec{B}##. In covariant formulation these come from the scalar ##F_{\mu \nu} F^{\mu \nu}## and the pseudoscalar ##\epsilon_{\mu \nu \rho \sigma} F^{\mu \nu} F^{\rho \sigma}##. For plane waves both invariants are 0.
