What is the significance of defining the trace of a tensor as $g^{\alpha\beta} R_{\alpha\beta}$ instead of $R_{\alpha\alpha}$ on a Riemannian manifold?
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The components of $\text{Ric}$ transform during coordinate change $x^\mu\mapsto \tilde{x}^\mu$ as $\tilde{R}_{\mu\nu}=\frac{\partial x^\sigma}{\partial \tilde{x}^\mu}\frac{\partial x^\rho}{\partial \tilde{x}^\nu}R_{\sigma\rho}$. This is just the usual transformation rule for coordinate-components of tensors.
Contracting over the two indices gives $$ \tilde{R}_{\mu\mu}=\frac{\partial x^\sigma}{\partial \tilde{x}^\mu}\frac{\partial x^\rho}{\partial \tilde{x}^\mu}R_{\sigma\rho}, $$ and there is no way to simplify this expression further, therefore this contraction is not coordinate-independent.
On the other hand for $R^\mu_{\ \ \nu}=g^{\mu\sigma}R_{\sigma\nu}$, the transformation rule is $$ \tilde{R}^\mu_{\ \ \nu}=\frac{\partial\tilde{x}^\mu}{\partial x^\sigma}\frac{\partial x^\rho}{\partial\tilde{x}^\nu}R^\sigma_{\ \ \rho}, $$ and contracting over this gives $$ \tilde{R}=\tilde{R}^\mu_{\ \ \mu}=\frac{\partial\tilde{x}^\mu}{\partial x^\sigma}\frac{\partial x^\rho}{\partial\tilde{x}^\mu}R^\sigma_{\ \ \rho}=\delta^\rho_\sigma R^\sigma_{\ \ \rho}=R^\sigma_{\ \ \sigma}=R. $$ Thatis why contractions may only be performed over mixed indices. The only thing the metric tensor does in that expression is to turn the two lower indices into one upper and one lower index.
Bence Racskó
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