$$
T(X, Y)=\nabla_{X} Y-\nabla_{Y} X-[X, Y],
$$
and thinking of the Riemann tensor as a map from three vector fields to a fourth one, we have (in funny-looking but standard notation)
$$
R(X, Y) Z=\nabla_{X} \nabla_{Y} Z-\nabla_{Y} \nabla_{X} Z-\nabla_{[X, Y]} Z .
$$
In these expressions, the notation $\nabla_{X}$ refers to the covariant derivative along the vector field $X$; in components, $\nabla_{X}=X^{\mu} \nabla_{\mu}$. So, for example, (3.116) is equivalent to
$$
\begin{aligned}
R_{\sigma_{\mu \nu}} X^{\mu} Y^{v} Z^{\sigma}=& X^{\lambda} \nabla_{\lambda}\left(Y^{n} \nabla_{\eta} Z^{\rho}\right)-Y^{\lambda} \nabla_{\lambda}\left(X^{\eta} \nabla_{\eta} Z^{\rho}\right) \
&-\left(X^{\lambda} \partial_{\lambda} Y^{\eta}-Y^{\lambda} \partial_{\lambda} X^{\eta}\right) \nabla_{\eta} Z^{\rho}
\end{aligned}
$$
6CCM334A COURSE NOTES ：
$$
p^{\lambda} \nabla_{\lambda} p^{\mu}=0
$$
By metric compatibility we are free to lower the index $\mu$, and then we may expand the covariant derivative to obtain
$$
p^{\lambda} \partial_{\lambda} p_{\mu}-\Gamma_{\lambda \mu}^{\sigma} p^{\lambda} p_{\sigma}=0 .
$$
The first term tells us how the momentum components change along the path,
$$
p^{\lambda} \partial_{\lambda} p_{\mu}=m \frac{d x^{\lambda}}{d \tau} \partial_{\lambda} p_{\mu}=m \frac{d p_{\mu}}{d \tau}
$$