Four-wavevector $k$

How do we define a four-vector?

Given two frames of reference, a four-vector $V$ is defined as a quantity which transforms according to the Lorentz transformation matrix $Lambda$:

begin{align} mathbf{V'} &= Lambda mathbf{V} tag{1} V^{mu} &= Lambda^{mu}_{nu}V^{nu} V_{mu} &= Lambda^{nu}_{mu}V_{nu} end{align}

How do we know $k_{mu}$ is a four-vector?

Because the phase is Lorentz invariant. It just means that $$phi equiv mathbf{k}cdotmathbf{r} - omega t = mathbf{k'}cdotmathbf{r'} - omega' t'. tag{2}$$

Why is $phi$ invariant?

Jackson gives the standard argument that the elapsed phase of the wave is proportional to the number of wave-crests that have passed the observer, and thus it must be frame-independent.

Also, interference effects are due to differences in the phase of waves. Notice that at a given space-time point, whether there is light or not cannot depend on the velocity of the observer looking there.

So we conclude that the phase $phi = mathbf{k}cdotmathbf{r} - omega t$ is a Lorentz invariant.

What are the components of $k_{mu}$?

You know the Lorentz transformations for spacetime vector $r = (mathbf{r}, ct)$. Apply them in eqn. $(2)$ to recover the forms for $k_{mu}$ which you have already mentioned in the question.

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Also, if you know that $p=(mathbf{p},E/c)$ forms a four-vector, and since $E =hbar omega$, $mathbf{p}=hbar mathbf{k}$, then $(mathbf{k}, omega/c)$ must be a four-vector too.

Is it enough to say that the phase:

$$ phi(x^{mu}) $$

is a scalar field, so that:

$$ k_{mu} equiv partial_{mu}phi(x^{mu})$$

is a 4-vector by manifest covariance?

Then, in any reference frame:

$$ partial_{mu}phi(x^{mu}) = (frac 1 c frac{partial phi}{dt}, vec nabla phi)=(frac{omega} c, vec k)$$