Diffeomorphism invariance of the Ricci tensor
Solution 1:
This question bothered me as well and later I found the solution, so I'm writing it here. It is known from basic Riemannian geometry that curvature is preserved by isometries. So if $\phi : (M,g) \to (\tilde{M}, \tilde{g})$ is an isometry, then $\phi^{*}R(\tilde{g}) = R(g)$.
But in our case, $\phi$ is just a diffeomorphism. But it is an isometry if considered as a map $\phi: (M, \phi^{*}g) \to (M,g)$.
Thus using isometry invariance of curvature we get that
$$ \phi^{*}\text{Ric}(g) = \text{Ric}(\phi^{*}g). $$