What is suggested in the question is essentially correct. When one defines some perturbative theory, one has to define which part of the Lagrangian is considered as a perturbation and which part is not. In the present case, the perturbation part is made of the terms 3,4,5,6. The Feynman rules for the "free propagators" always appear as inverse of the corresponding terms in the Lagrangian whereas the Feynman rules for the "interactions" always appear as the corresponding terms in the Lagrangian. The easiest way I find to remember this is the path integral derivation of the Feynman rules. Here is a finite dimensional analogue: to compute perturbatively the integral $\int_{\mathbb{R}}dx e^{-\frac{1}{2} ax^2 + V(x)}$ where $V(x)$ is the perturbation simply means to expand $e^{V(x)}$ and to integrate each term of the expansion against the gaussian $e^{-\frac{1}{2}ax^2}$. It is obvious that the terms in $V(x)$ will appear to some positive power whereas $a$ will appear with some negative power because

$\int_{\mathbb{R}} dx e^{-\frac{1}{2}a x^2}x^{2n} = constant \frac{1}{a^n}$.