Intuition for the Ricci tensor of a Reissner-Nordström black hole

Actually, the Reissner-Nordström metric has $$h(r)=1-2m/r+q^2/r^2tag1.$$ (The charge is squared.)

The Ricci tensor is diagonal with $$R_{tt}=-frac{q^2}{r^4}g_{tt}tag{2a},$$ $$R_{rr}=-frac{q^2}{r^4}g_{rr}tag{2b},$$ $$R_{thetatheta}=frac{q^2}{r^4}g_{thetatheta}tag{2c},$$ $$R_{phiphi}=frac{q^2}{r^4}g_{phiphi}tag{2d}.$$

This can be understood based on the form of the energy-momentum tensor $T_{munu}$. It is purely electromagnetic, with

$$4pi T_{munu}=F_{mualpha}F_nu{}^alpha-frac14g_{munu}F^2tag{3}$$

where $F_{munu}$ is the electromagnetic field tensor due to the hole’s charge $q$.

The electromagnetic field of a Reissner-Nordström hole consists of a radial electric field and no magnetic field. Its nonzero tensor components are only $F_{tr}$ and $F_{rt}$. In this case one finds that the nonzero components of (3) are

$$4pi T_{tt}=frac14g_{tt}F^2tag{4a},$$ $$4pi T_{rr}=frac14g_{rr}F^2tag{4b},$$ $$4pi T_{thetatheta}=-frac14g_{thetatheta}F^2tag{4c},$$ $$4pi T_{phiphi}=-frac14g_{phiphi}F^2tag{4d}$$

where

$$F^2=F_{munu}F^{munu}=-2frac{q^2}{r^4}tag{5}.$$

Finally, the Ricci scalar $R$ vanishes (which happens because the electromagnetic energy-momentum tensor is traceless), so the Einstein field equations simplify to

$$R_{munu}=8pi T_{munu}tag{6}.$$

Combining (6), (4), and (5) gives (2).

So the short answer is that the relationship you found is because the electric field is radial and the magnetic field is zero. Such an electromagnetic field has a mixed energy-momentum tensor of the form $T^mu{}_nuproptotext{diag}(-1,-1,1,1)$ in spherical coordinates $(t,r,theta,phi)$, and so the mixed Ricci tensor has the same form.