= Solution
An affinely parametrized <geodesic> satisfies $d^2x^a/d\lambda^2+\Gamma^a{}_{bc}T^bT^c=0$, which is exactly $\nabla_TT=0$. The <Levi-Civita connection> is metric-compatible, so
$$
\frac d{d\lambda}g(X,Y)=(\nabla_Tg)(X,Y)+g(\nabla_TX,Y)+g(X,\nabla_TY)=0
$$
when both vectors are parallel transported. \b[Their inner product is constant], including their lengths; this preservation in fact holds along any path with parallel vectors, not only a <geodesic>.
For the two-parameter family, mixed partial derivatives commute and $\Gamma^a{}_{bc}$ is symmetric in its lower indices. Consequently
$$
\nabla_TS^a=\partial_\lambda\partial_\mu x^a+\Gamma^a{}_{bc}S^bT^c=\nabla_ST^a.
$$
Using this equality and the supplied <curvature>-commutator convention,
$$
\nabla_T^2S=\nabla_T\nabla_ST=\nabla_S\nabla_TT+[\nabla_T,\nabla_S]T=R(T,S)T.
$$
In components the <geodesic deviation> equation is
$$
\boxed{\nabla_T^2S^a=R^a{}_{bcd}T^bT^cS^d.}
$$
It describes the relative acceleration of nearby freely falling worldlines. A freely falling observer can remove the connection at an event by choosing a local inertial frame, but cannot remove the <curvature>-dependent tidal acceleration of neighboring freely falling particles. Measuring their separation and relative acceleration therefore measures the local gravitational tidal field.
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