Most statements are clear, except for the last one. To see the last identity, fix \(a_{1},\ldots,a_{n}\in\mathbb{R}\) and set
\begin{equation} v:=a_{1}\frac{\partial}{\partial x^{1}}\bigg{|}_{p}+\ldots+a_{n}\frac{\partial}{\partial x^{n}}\bigg{|}_{p}.\nonumber \\ \end{equation}Then
\begin{equation} \gamma^{\prime}_{v}(t)=a_{1}\frac{\partial}{\partial x^{1}}\bigg{|}_{\gamma_{v}(t)}+\ldots+a_{n}\frac{\partial}{\partial x^{n}}\bigg{|}_{\gamma_{v}(t)}.\nonumber \\ \end{equation}By the geodesic equation, we find that for any \(t\) close to \(0\) we have
\begin{equation} 0=\frac{D}{dt}\gamma^{\prime}_{v}(t)=\frac{D}{dt}\Big{(}a_{1}\frac{\partial}{\partial x^{1}}\bigg{|}_{\gamma_{v}(t)}+\ldots+a_{n}\frac{\partial}{\partial x^{n}}\bigg{|}_{\gamma_{v}(t)}\Big{)}=\sum_{i,j=1}^{n}a_{i}a_{j}\nabla_{\partial/\partial x^{i}}\frac{\partial}{\partial x^{j}}\bigg{|}_{p}.\nonumber \\ \end{equation}Define
\begin{equation} Q((a_{1},\ldots,a_{n}),(b_{1},\ldots,b_{n}))=\sum_{i,j=1}^{n}a_{i}b_{j}\nabla_{\partial/\partial x^{i}}\frac{\partial}{\partial x^{j}}\bigg{|}_{p}.\nonumber \\ \end{equation}By torsion-freeness, \(Q\) is a symmetric bilinear form and
\begin{equation} Q((a_{1},\ldots,a_{n}),(a_{1},\ldots,a_{n}))=0.\nonumber \\ \end{equation}So \(Q\equiv 0\). ∎