You don't need the somewhat cumbersome velocity additions (at least not for free em. fields). You just have to consider that the phase of a plane-wave solution, ##\omega/c t-\vec{k} \cdot \vec{x}=k_{\mu} x^{\mu}## is a Lorentz scalar, which implies that ##k_{\mu}## is a Lorentz vector. So you can just using a Lorentz boost to transform the components of the wave-fourvector ##k^{\mu}## wrt. an inertial frame to any other inertial frame.
If you are only interested in the change of the frequency of the light as seen from different observers, you only need to know that it can only depend on the relative velocity between the light source and the observer. So let ##u^{\mu}=\gamma(1,\vec{\beta})## be the four-velocity (normalized to 1) of the light source in the reference frame of the observer. Now it's most convenient to refer to the frequency of the light ##\omega_0## in the restframe of the light source. It is a Lorentz invariant and given by
$$\omega_0/c=u_{\mu} k^{\mu}=\gamma (\omega/c -\vec{\beta} \cdot \vec{k}).$$
Now with ##\vec{k}=k \vec{n}## and the dispersion relation ##k=\omega/c## you get
$$\omega_0=\gamma \omega (1-\vec{\beta} \cdot \vec{n})$$
or
$$\omega=\frac{\omega_0}{\gamma (1-\vec{\beta} \cdot \vec{n})}.$$
Now take the extreme cases: ##\vec{\beta}=\beta \vec{n}##, i.e., the light-source travels in the diretion of wave propagation (i.e., towards the observer). This gives ##\gamma(1-\beta)=\sqrt{(1-\beta)/(1+\beta)}## and thus
$$\omega_{\text{max blue}}=\sqrt{\frac{1+\beta}{1-\beta}} \omega_0.$$
This is the maximal blue shift you can get compared to the frequency in the rest frame of the light source. For the maximal redshift you have to set ##\vec{\beta}=-\beta \vec{n}## (source moving away from observer):
$$\omega_{\text{max red}}=\sqrt{\frac{1-\beta}{1+\beta}} \omega_0.$$
Then a specifically relativistic effect is that there's also a shift if the source moves perpendicular to the light-propagation direction, i.e., for ##\vec{\beta} \cdot \vec{n}=0##:
$$\omega_{\perp}=\frac{\omega_0}{\gamma}=\sqrt{1-\beta^2} \omega_0,$$
i.e., you have a red shift, which is of course purely due to time dilation between the light-source restframe and the observer's frame. That's the transverse Doppler effect.