Lemma: Palatini form of the first variation

For a smooth compactly supported variation $\delta q_{\mu \nu }$,

Consequently

Proof. $\delta {{\Gamma }^{\rho }}_{\mu \nu }$ is a tensor. Both sides of $\delta R^{\rho }{}_{\sigma \mu \nu }={\nabla }_{\mu }\delta {{\Gamma }^{\rho }}_{\nu \sigma }-{\nabla }_{\nu }\delta {{\Gamma }^{\rho }}_{\mu \sigma }$ are tensors, and the identity holds in normal coordinates at any point, where $\Gamma =0$ and $\nabla =\partial$. Contracting, $\delta R_{\sigma \nu }={\nabla }_{\rho }\delta {{\Gamma }^{\rho }}_{\nu \sigma }-{\nabla }_{\nu }\delta {{\Gamma }^{\rho }}_{\rho \sigma }$, so $q^{\sigma \nu }\delta R_{\sigma \nu }={\nabla }_{\rho }{W}^{\rho }$ and $\sqrt{-q}{\nabla }_{\rho }{W}^{\rho }={\partial }_{\rho }(\sqrt{-q}{W}^{\rho })$. With $\delta \sqrt{-q}=\frac{1}{2}\sqrt{-q}{q}^{\mu \nu }\delta q_{\mu \nu }$ and $\delta q^{\mu \nu }=-q^{\mu \alpha }{q}^{\nu \beta }\delta q_{\alpha \beta }$,

The chain rule $\delta q=\kappa \delta h={\kappa }^2\delta k$ gives the remaining formulas.