In an exercise for a course on quantum field theory, I am given the following Lagrangian: $$ \mathcal{L} = -\frac{1}{2} G_{\mu\nu}^a G^{a\mu\nu} + 2 (D_\mu \phi^\dagger)^a(D^\mu \phi)^a - 2 m^2\phi^{\dagger a}\phi^a -2\lambda (\phi^{\dagger a}\phi^a)^2, $$ where the covariant derivative and the strength tensor are given as $$ G_{\mu\nu} = \partial_\mu G_\nu - \partial_\nu G_\mu - i g [G_\mu, G_\nu], $$ $$ D_\mu = \partial_\mu - i g G_\mu \equiv \partial_\mu - K_\mu, $$ and where the scalar field is complex and decomposes under the fundamental representation of ${\rm SU}(N)$ as $\phi = \phi^a T^a$, for some given $N\geq 2$. Of course, repeated indices are summed over.
Now the first question I am asked is to calculate the propagator for the scalar field, and here a conceptual problem that has been haunting me for a while makes its appearance again: when we talk about the propagator of a given field for a specific Lagrangian, aren't we talking about the Green's function for the differential equation that gives the dynamics of the (classical) field? Applying calculus of variation to the action corresponding to this Lagrangian under a change $\phi\mapsto \phi+\delta \phi$ I reach the equations of motion $$ \Omega^{ab} \phi^b = \Omega^{ab} \phi^{\dagger b} = 0 \tag{1} $$
for an operator
$$ \Omega^{ab} = \partial_\mu K^{\mu a b} - K_\mu^{ab} \partial^\mu + K^{ca}_\mu K^{\mu cb} - (\partial^2 +m^2)\delta^{ab} $$
where, as usual, inside a linear operator terms like $\partial_\mu A$ act as $\phi\mapsto \partial_\mu(A \phi)$. Now the provided solution to the exercise says
The propagator for the scalar field is contained in the kinetic term $$\mathcal{L}_{\rm kin} = -\phi^{\dagger a}(\partial^2 + m^2)\delta^{ab}\phi^b \tag{2}$$
so according to the solution the propagator is actually the Green function for the differential equation induced by $(2)$ and not by $(1)$, but why? Extremizing the action under a variation of $\phi$ for the full Lagrangian leads to (up to possible mistakes in my development) equation $(1)$ and as such I would expect to be that what determines the propagator. Is it because the (free) propagator we are looking for here derives from the interaction-free part of the Lagrangian, which by definition only includes (for $\phi$) the last term in my $\Omega^{ab}$?
If this is the reason, I would still appreciate further information on why the propagator doesn't include the whole $\Omega$. I understand that the "practical" objective is to express through the propagator $\Delta$ a free partition function as products of terms like $$ Z_{\rm free}\propto \exp\left[i\int d^4xd^4y J(x)\Delta(x-y)J(y)\right] \qquad (3) $$ but, to my intuition, the interactions between a field and a gauge field are of different "nature" than the interactions added by hand such as the $\lambda$ term, because the former appear by the necessity of imposing gauge invariance. However, I believe these other interactions "added by hand" are also incorporated by a necessity, namely making the theory renormalizable, so perhaps they are not that different in nature after all, are they?