Feynman rules Horndeski theory Let $\phi$ be a scalar field and $g_{\mu \nu} = \eta_{\mu \nu}+h_{\mu \nu}/M_p$ where $M_p$ is the Planck mass (so we assume we deal with perturbations). Let $\Lambda_2,\Lambda_3$ be energy scales such that $\Lambda_2 \gg \Lambda_3$. These are defined by $\Lambda^2_2 = M_p H_0$ and $\Lambda_3^3 = M_p H_0^2$. The Horndeski action is:
$$S = \int d^4 x \sqrt{-g} \sum^5_{i=2} \mathcal{L}_i,$$
where
\begin{align}
\mathcal{L_2}&=\Lambda_2^4 G_2,\nonumber \\
\mathcal{L_3}&=\Lambda_2^4 G_3 [\Phi], \nonumber \\
\mathcal{L_4}&= M_p^2 G_4 R + \Lambda_2^4 G_{4,X}([\Phi]^2 - [\Phi^2]),\nonumber \\
\mathcal{L_5}&= M_p^2 G_5 G_{\mu \nu}\Phi^{\mu \nu} - \frac{1}{6}\Lambda_2^4 G_{5,X}([\Phi]^3 - 3[\Phi][\Phi^2] + 2[\Phi^3]), \nonumber
\end{align}
where $G_2,G_3,G_4,G_5$ are functions of $\phi$ and $X = -\frac{1}{2}\nabla^\mu \phi \nabla_\mu \phi /\Lambda_2^4$, $\Phi^{\mu}_{ \ \nu}:= \nabla^\mu \nabla_\nu \phi/\Lambda_3^3$ and square brackets indicate the trace, e.g. $[\Phi^2] = \nabla^\mu \nabla_\nu \phi \nabla^\nu \nabla_\mu \phi/\Lambda_3^6$ and $,$ denote partial derivatives.
In the paper arXiv:1904.05874 they state the Feynman rules in the Appendix. In the paper they try to derive the Feynman rules from the above action around the vacuum expectation value $\langle \phi \rangle = 0$. It says for instance that the leading-order vertex for $\phi\phi\phi$ is given by:
$$-\frac{\bar{G}_{3,X} + 3\bar{G}_{4,\phi X}}{3\Lambda_3^3}\delta^{\mu \nu}_{\alpha \beta}\phi \phi^\nu_\mu \phi^\beta_\alpha,$$
where the bar indicates that the function is evaluated at $\langle \phi \rangle = 0$, $\delta$ is the generalized Kronecker delta and $\phi^\nu_\mu:= \nabla^\nu \nabla_\mu \phi$.
My question is how can I derive such an expression from the action?
 A: I'm just going to summarize some discussion in the comments, since comments get deleted and it ended up building up to half an answer.
Let's take a simple example, $G=-\frac{1}{2} a \phi (\partial \phi)^2 = a \phi X$, where $X\equiv -\frac{1}{2} (\partial \phi)^2$. Then we perturb $\phi$
\begin{equation}
\phi = \bar\Phi + \varphi
\end{equation}
Even though we will ultimately set $\langle \phi \rangle = \bar\Phi = 0$, we do not want to do this yet. First, we expand the Lagrangian; in our example this is $G$
\begin{equation}
G = a \left(\bar\Phi \bar X - \bar\Phi \partial_\mu \bar \Phi \partial^\mu \varphi + \bar\Phi Y + \varphi \bar X - \varphi \partial_\mu \bar \Phi \partial^\mu \varphi + \varphi Y \right)
\end{equation}
where I've defined $\bar{X} = - \frac{1}{2} (\partial \bar\Phi)^2 $ and $Y = -\frac{1}{2} (\partial \varphi)^2$.
Now we set $\bar\Phi = \bar X = 0$. Then our example reduces to
\begin{equation}
G = a \varphi Y
\end{equation}
Since $G_{\phi X}(\phi=\bar\Phi=0, X=\bar X=0) = a$, we can equivalently write this as
\begin{equation}
G = \bar{G}_{\phi X} \varphi Y
\end{equation}
To get the Feynman rules from the Lagrangian for $\varphi$, one then follows one of the normal procedures, for example differentiating the partition function, as described in many sources. An example source that is free is Chapter 10 of Srednicki's lecture notes of QFT (and the preceding chapters which set the stage for chapter 10): cns.gatech.edu/FieldTheory/extras/SrednickiQFT03.pdf. In the case of this example, there will be a cubic vertex with an associated factor of something like $i \bar{G}_{\phi X} k^2$ (I am not 100% sure about the factor of $i$... I think you get a $-i$ from the definition of the amplitude, $-1/2$ from the definition of $Y$, and $2$ from the fact that the two legs with $\partial \phi$ are identical... but this should be checked carefully)
