Why do we introduce the wave vector $k$ into the wave solution? If every function of the form $f(x-vt)$, in the one dimensional case for example, solves the wave equation for a wave propagating in the positive $x$ direction, why do we always represent the solution as $f(kx-\omega t - \varphi)$? How and why would we go from the first solution to the other? In textbooks they just search for a solution of this form and then get the requirement $k = \frac{\omega}{v}$ but I lack any intuition for even looking at this kind of solution.
 A: The expression f(x – vt)  indicates that you have a function moving in the positive x direction with a speed v.  But many waves can be represented by sine or cosine functions (or combinations of these) which require angles (usually in radians) as their arguments.  The k  and ω put the arguments in radians and also introduce properties of the wave (frequency and wavelength) into the function. The phase constant indicates which part of the wave is at x = 0 when t = 0.
A: Your question:

In textbooks they just search for a solution of this form and then get the requirement $k=\frac{\omega}{v}$ but I lack any intuition for even looking at this kind of solution.

I think you would have the intuition to look for that kind of solution.  You probably know that $\lambda f=v$.  Well, this should give you a hint, such as $2\pi f = \omega$ that you can use in the following way:
$$
x-vt = x-\lambda\,f\,t=x - \lambda \,\frac{\omega}{2\pi} \,t = \frac{2\pi}{\lambda}\,x -\omega\,t 
$$
And, since you probably know that the definition of $k$ is merely $\frac{2\pi}{\lambda}$ and making this substitution you arrive at: $x-vt=k x - w t$ and the phase angle $\phi$ merely makes it pretty.
A: The meaning of $\mathbf k$ is related to a plane wave travelling in an arbitrary direction. $\mathbf k$ defines the direction of the wave.
$f(\mathbf {k.x} - \omega t - \phi)$
where $\omega = |\mathbf k|v$
$\mathbf {k.x} = k_xx + k_yy + k_zz$. For each instant $t_0$,  $k_xx + k_yy + k_zz = \omega t_0 + \phi$ is an equation of a plane. So the function has the same value on this plane.
$f$ is an arbitrary function, not necessarily a periodic, not even an oscillatory function to be a solution of the wave equation.
For a frame moving in the $k$ direction and with a velocity $v$, there is a static field, where the planes normal to $k$ are the level curves and the gradient is always parallel to $k$.
