Why is the Ricci tensor defined as $R^\mu _{\nu \mu \sigma}$?

There are only six possible contractions, each of which can be simplified using the symmetries of the Riemann tensor:

$R^{\mu}_{\enspace\mu\lambda\sigma}=0$ because $R_{\kappa\nu\lambda\sigma}=-R_{\nu\kappa\lambda\sigma}$.

$R^{\mu}_{\enspace\nu\mu\sigma}=\text{Ric}_{\nu\sigma}$ is the usual definition.

$R^{\mu}_{\enspace\nu\lambda\mu}=-\text{Ric}_{\nu\lambda}$ because $R_{\kappa\nu\lambda\sigma}=-R_{\kappa\nu\sigma\lambda}$.

$R^{\enspace\mu}_{\kappa\enspace\mu\sigma}=-\text{Ric}_{\kappa\lambda}$ because $R_{\kappa\nu\lambda\sigma}=R_{\nu\kappa\lambda\sigma}$.

$R^{\enspace\mu}_{\kappa\enspace\lambda\mu}=\text{Ric}_{\kappa\lambda}$ because $R_{\kappa\nu\lambda\sigma}=R_{\nu\kappa\sigma\lambda}$.

$R^{\enspace\enspace\mu}_{\kappa\lambda\enspace\mu}=0$ because $R_{\kappa\nu\lambda\sigma}=R_{\kappa\nu\sigma\lambda}$.

So all the possible contractions result either in the Ricci tensor, its negation, or zero. Thus the Ricci tensor is unique: it (or its negation) is the only non-zero order 2 tensor you can make by contracting the Riemann tensor.

The Ricci tensor's geometric interpretation is that it describes how open balls on a manifold behave i.e. how the radius of an open ball behaves. This is naturally a symmetric quantity since the manifold is constructed in a way in which you have a freedom of choice of what basis to choose. The symmetries of the Riemann tensor are $$R_{abcd} = R_{cdab} = - R_{cdba} = R_{cdba}.$$ SO it is obvious from the symmetric properties that you have to contract either the first and third or second and fourth indices with each other. And now you choose to define a symmetric contraction of the Riemann tensor as $$R_{bd} = R_{db} = R^a_{\ bad}. $$

It's a choice based on symmetry.