The \(n\)-th covariant derivative of \(F\) can be expressed as
\begin{equation*} \nabla_{X_n,\ldots ,X_1}^n F = \tr(\nabla^{n-1}_{X_n,\ldots ,X_2}(\nabla F)\otimes X_1). \end{equation*}This can be verified by comparing the component formulas.
We prove the result by induction. For \(n=1\) the formula is true. Assume the formula is true for some \(n\in \mathbb{N}\) and denote some extension of \(\gamma'\) by \(\widetilde{\gamma}'\). Then
\begin{align*} \nabla_{\gamma'}(\nabla^{n}_{\gamma'} F) &= \nabla_{\gamma'}\bigl(\tr(\nabla^{n-1}_{\gamma'}(\nabla F) \otimes \widetilde{\gamma}')\bigr)\\ &= \tr(\nabla_{\gamma'}(\nabla^{n-1}_{\gamma'}(\nabla F) \otimes \widetilde{\gamma}')) \\ &= \tr(\nabla_{\gamma'}(\nabla^{n-1}_{\gamma'}(\nabla F)) \otimes \widetilde{\gamma}') + \tr(\nabla^{n-1}_{\gamma'}(\nabla F) \otimes \nabla_{\gamma'}\widetilde{\gamma}')\\ &= \tr(\nabla^{n}_{\gamma'}(\nabla F)\otimes \gamma') \\ &= \nabla^{n+1}_{\gamma'} F \end{align*}where we sequentially used the equation above, the fact that the covariant derivative commutes with the trace operator, the product rule for tensor products, the induction assumption and \(\nabla_{\gamma'}\widetilde{\gamma}'\equiv 0\) since \(\gamma\) is geodesic.