A radial geodesic in normal coordinates is given by $\gamma:t \mapsto t(V_1,....,V_n).$ Is it then true that any normal coordinate $\partial_x|_{\gamma}$ is parallel along $\gamma,$ i.e. $\nabla_{\gamma'}\partial_x=0$? I implicitly saw this once in a calculation, but I have never seen this anywhere as a property of normal coordinates, so I guess it is false, but I don't really know.
If anything is unclear, please let me know.
No, it's not true. To see why, choose an index $i$ and define a vector field $J$ along $\gamma$ by $$ J(t) = t\partial_{x^i}\big|_{\gamma(t)}. $$ Then $J$ is the variation field of the following family of geodesics: $$ \gamma_s(t) = (tV_1,\dots,t(V_i+s),\dots,tV_n). $$ Therefore, $J$ satisfies the Jacobi equation: $$ \nabla_{\gamma'}\nabla_{\gamma'} J(t) = R(J(t),\gamma'(t))\gamma'(t), $$ whose right-hand side is not necessarily zero unless the metric is flat. If $\partial_{x^i}$ were parallel along $\gamma$, then $J$ would satisfy $$ \nabla_{\gamma'}\nabla_{\gamma'} J(t) = \nabla_{\gamma'} \left(\partial_{x^i} + t \nabla_{\gamma'} \partial_{x^i}\right) = 2\nabla_{\gamma'} \partial_{x^i} + t \nabla_{\gamma'}\nabla_{\gamma'} \partial_{x^i}=0. $$