Gauge dependence of Feynman propagator

In QED we have the property that any gauge dependence must cancel out in physical quantities. Nevertheless Feynman rules display some gauge dependence. This is what I am a little unclear on.

The obvious place gauge dependence comes into Feynman rules is through the propagator
$$frac{-i}{k^2 – m^2} bigg( eta^{mu nu} – (1-xi)frac{k^{mu}k^{nu}}{k^2} bigg) $$
I take it that if I calculated some physical quantity using this general form of the propagator there would be no $xi$ dependence at the end of the day. However we also have freedom in the polarisation vectors. Under a gauge transformation $$epsilon^mu_r longrightarrow epsilon^mu_r + i k^mu c_r(k)$$ the amplitude should be invariant. For external photons $mathcal{M} = epsilon^mu_r mathcal{M}^mu $ then this works nicely via the Ward identity $k_mu mathcal{M}^mu = 0$.

On the other hand deriving the photon propagator in a slightly more physical way (rather than via gauge fixing) we find that the numerator of the propagator actually arises from the polarisation sum
$$ sum_r epsilon^mu_r epsilon^{nu *}_r $$
I therefore expect the polarisation sum to change as you change the gauge. Does this work out the way I expect? Do we have that:

1. In the Feynman gauge ($xi = 1)$ the polarisation sum is $sum_r epsilon^mu_r epsilon^{nu *}_r = -eta^{mu nu}$.
2. If I then do $epsilon^mu_r longrightarrow epsilon^mu_r + i k^mu c_r(k)$ then the polarisation sum is $sum_r ( epsilon^mu_r + i k^mu c_r(k)) ( epsilon^{nu}_r + i k^nu c_r(k))^* $ which allows me to relate the new $xi$ and the gauge transformation function $c_r(k)$? Exactly how does this work? Aren’t I going to get cross terms $ ~ k^mu epsilon^{nu}_r $?

Many thanks in advance.