= Solution
Although <connection coefficients> are not <tensor> components, their variation $C^a{}_{bc}=\delta\Gamma^a{}_{bc}$ is <tensorial>: the <difference of affine connections is a tensor>. Varying the <Levi-Civita connection> at fixed coordinates gives
$$
C^a{}_{bc}=\frac12g^{ad}
(\nabla_b\delta g_{cd}+\nabla_c\delta g_{bd}-\nabla_d\delta g_{bc}).
$$
In the paper's <curvature sign convention>, varying the <derivative> and quadratic-connection terms and grouping them into <covariant derivatives> gives the <Palatini identity>
$$
\boxed{\delta R_{bc}=\nabla_cC^a{}_{ba}-\nabla_aC^a{}_{bc}.}
$$
One may verify the grouping in <normal coordinates> at a point, where the <affine connection> vanishes and the expression is just the difference of two <partial derivatives>; both sides are <tensors>, so it holds in every <coordinate frame>. Reversing the definition of the <Riemann curvature tensor> reverses this identity's right-hand side.
For completeness, put $q_{ab}=\delta g_{ab}$ and $q=g^{ab}q_{ab}$. The contracted variation uses $C^a{}_{ba}=\nabla_bq/2$ and $g^{bc}C^a{}_{bc}=\nabla_bq^{ab}-\nabla^aq/2$. After two <integrations by parts>,
$$
\int\sqrt{-g}\,\Phi g^{bc}\delta R_{bc}\,d^4x
=\int\sqrt{-g}\left[-\nabla_a\nabla_b\Phi\,q^{ab}
+\Box\Phi\,q\right]d^4x,
$$
up to the discarded <boundary term>. Substituting $q^{ab}=-\delta g^{ab}$ gives the differentiated-scalar terms in the <metric variation of a scalar-curvature coupling>.
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