There is no need to consider photons, when a plane-wave will do. A place wave is not a photon, nor is a photon a plane wave. What they have in common is that they can be defined by an exact frequency ($\omega$) or wave-vector ($\vec k$), with $\omega = |k|c$.
Either-way, your statement is correct: charge (and current) distributions don't affect $\vec k$ because the solutions superimpose.
Now regarding charge particle and force:
"how do charged particles exert force on each other instead of point values in the EM field just superimposing?"
It is hard to make sense of the question. If we have $q_1$ and $q_2$, each contributes a field $\vec E_1(\vec r)$ and $\vec E_2(\vec r)$ for a total field:
$$ \vec E(\vec r) = \vec E_1(\vec r) + \vec E_2(\vec r) $$
So is it that you expect the force on $q_1$ to be:
$$ \vec F_1 = q_1 \vec E(\vec r_1) $$
rather than:
$$ \vec F_1 = q_1 \vec E_2(\vec r_1) $$
Thereby missing a term:
$$ q_1 \vec E_1(\vec r_1) $$
The thing is, $\vec E_1(\vec r_1)$ is not defined. All we know is that
$$ (\nabla\cdot\vec E_1)(\vec r_1) = \frac {4\pi} {\epsilon_0}\delta(\vec r-\vec r_1)$$
So we know the divergence is infinite, and isotropic. There is really no way to get a force (with a direction) out of that.
Moreover, the definition of the electric field is a the force per unit charge on a fiducial charge. Since EM is linear, the fiducial charge can be of any magnitude.
Such reasoning does work in gravity, nor QCD, which are both nonlinear.
