I have two definitions of the Riemann tensor which are $$\Large R^{\rho}_{\,\,\,\sigma\mu\nu}=\partial_{\mu}\Gamma^{\rho}_{\nu\sigma}-\partial_{\nu}\Gamma^{\rho}_{\mu\sigma}+\Gamma^{\rho}_{\mu\lambda}\Gamma^{\lambda}_{\nu\sigma}-\Gamma^{\rho}_{\nu\lambda}\Gamma^{\lambda}_{\mu\sigma}=R_{\nu\mu\sigma}^{\,\,\,\,\,\,\,\,\,\rho}$$
But can someone demonstrate how these two ways of writing the Riemann tensor are equivalent?
The fact follows from the two identites of the Riemann tensor: $R_{ijkl}=R_{klij}=-R_{lkij}=R_{lkji}$. Now, $R_{jkl}^i =g^{im} R_{mjkl}$ and ${R_{jkl }}^i=g^{im}R_{jklm }$.
With that you should be able to get the result.
You can use the Riemann if you want. However, for 3D spaces, the Ricci is enough. Anyway, given the metric, remember that the expression for the Christoffel symbols is $$\Gamma_{ijk}=\frac{1}{2}\left( \frac{\partial g_{ij}}{\partial x^k} + \frac{\partial g_{ik}}{\partial x^j} - \frac{\partial g_{jk}}{\partial x^i} \right) .$$ Moreover, your metric is diagonal ($g=diag(u^2+v^2,u^2+v^2,u^2v^2$). Hence some of the Christoffel symbols are $0$.
By the way, your space seems a 2D surface. In this case, according to Wiki, the Riemann is given in coordinates just by $$R_{ijkl}= K(g_{ik}g_{kl}-g_{il}g_{jk}),$$ where $K$ is the Gaussian curvature