In the context of Lagrangian field theories I am reading Quantum Fields and Strings for the mathematician where in chapter 3.2 (page 192) the authors set up the Lagrangian for a real scalar field $\Phi : M \to \mathbb{R}$
\begin{equation} L(\Phi) = \frac{1}{2} ( d\Phi \wedge \star d\Phi - m^2 \Phi \wedge \star \Phi) . \end{equation} Here $d$ denotes the differential on the spacetime manifold $M$ and $\star$ the Hodge star. Denoting $\delta$ the "variational" differential on $\cal{F}$ (the space of fields) they then compute (using the notation $\phi : \cal{F} \times M \to \mathbb{R}, (\Phi, x) \mapsto \Phi(x) $ )
\begin{equation}
\delta L = - \delta \phi \wedge (d \star d \phi + m^2 \star \phi) - d(\delta \phi \wedge \star d \phi)
\end{equation}
But I can't quite get there reasonably. My try with "formal manipulations" led to :
\begin{equation}
\delta(d \phi \wedge \star d \phi) = \delta d \phi \wedge \star d \phi + (-1)^a d\phi \wedge \delta \star d \phi = d\delta \phi \wedge [-2 \star d\phi ]
\end{equation}
where I used $d \delta = -\delta d $ like in Anticommutation of variation $\delta$ and differential $d$ and that $\delta \star = \star \delta$ as well as $w \wedge \star \eta = \eta \wedge \star w$. Here for the graded Leibniz rule I 'chose' $a=0$ because otherwise "it wouldn't work". On the other hand, for reexpressing the last expression in the previous equation,
\begin{equation}
d(\delta \phi \wedge \star d \phi) = d\delta\phi \wedge \star d \phi + (-1)^b \delta \phi \wedge d \star d \phi
\end{equation}
but this time $b=1$ is necessary for it to yield the correct $\delta L$ from above.
Question In the above cases i could not naively apply the graded Leibniz using that the degree of the forms $d\phi$ and $\delta \phi$ both seem to be 1, so what is the correct rule for determining expressions such as $d( w \wedge \eta), \delta(w \wedge \eta)$ etc. or also for $$\mathrm{d} = d + \delta ~ ? $$
