I use an automated program that shields me from the details of the claculations, which are long and error-prone.
But one can write
$$R^\rho{}_{\sigma\mu\nu} = \partial_\mu\Gamma^\rho{}_{\nu\sigma}
- \partial_\nu\Gamma^\rho{}_{\mu\sigma}
+ \Gamma^\rho{}_{\mu\lambda}\Gamma^\lambda{}_{\nu\sigma}
- \Gamma^\rho{}_{\nu\lambda}\Gamma^\lambda{}_{\mu\sigma}$$
to find the Riemann, and the Ricci is then the contraction of the first and third slots of the Riemann, i.e.
$$R_{\mu\nu} = R^{\rho}{}_{\mu \rho \nu}$$
See
https://en.wikipedia.org/wiki/Riemann_curvature_tensor
You'll need to be familiar with the Einstein summation convention
https://en.wikipedia.org/wiki/Einstein_notation and the concept of tensor contraction,
https://en.wikipedia.org/wiki/Tensor_contraction
If you're not already familiar with these concepts, I'm not sure Wiki will explain it in enough detail for you to figure it out without a textbook (and a lot of study), but at least you'll have some idea of what to look for.
Also, I've not written out all the permutations of terms that are given by symmetries in the above. Using my automated program to calculate exactly what I wrote above, I get:
$$R^r{}_{\phi r \phi} = -R^r{}_{\phi\phi r} = kr^2$$
$$R^\phi{}_{r r \phi} = -R^\phi{}_{r \phi r} = \frac{k}{kr^2-1}$$
I tried not to make any typos, but I wouldn't guarantee it absolutely.
