= Solution
Write $D_t=\nabla_{\dot\gamma}$ and let $P_t:T_pM\to T_{\gamma(t)}M$ denote <parallel transport>. Since $\gamma$ is a <geodesic>, $D_t\dot\gamma=0$, while the transported vectors satisfy $D_te_i(t)=0$. The <locally symmetric Riemannian manifold> condition is $\nabla R=0$. Hence the <Levi-Civita connection> product rule yields
$$
D_t\bigl(R(\dot\gamma,e_i(t))\dot\gamma\bigr)
=(\nabla_{\dot\gamma}R)(\dot\gamma,e_i(t))\dot\gamma
+R(D_t\dot\gamma,e_i(t))\dot\gamma
+R(\dot\gamma,D_te_i(t))\dot\gamma
+R(\dot\gamma,e_i(t))D_t\dot\gamma=0.
$$
Both this vector field and $\lambda_i e_i(t)$ are parallel and agree at zero. Uniqueness of <parallel transport> therefore proves
$$
\boxed{K_{\dot\gamma(t)}e_i(t)=\lambda_i e_i(t),\qquad K_{\dot\gamma(t)}=P_tK_vP_t^{-1}.}
$$
The <eigenvalues> are independent of $t$. The original PDF has $K_{\dot\gamma(t)}$ here; the converted TeX omits the dot, which would incorrectly put a point rather than a tangent vector in the operator's subscript.
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