Let $(M,\langle\cdot,\cdot\rangle)$ be a pseudo-Riemannian manifold and $\nabla$ its Levi-Civita connection. If one defines the connection $\overline{\nabla}$ in $TM\oplus {\rm End}(TM)$ by $$\overline{\nabla}_X(Y,F) = (\nabla_XY- F(X), \nabla_XF - R(X,Y)),$$then for any Killing field $\xi$ on $M$, we have that $(\xi, \nabla\xi)$ is a $\overline{\nabla}$-parallel section of $TM\oplus {\rm End}(TM)$, and in particular this gives a proof that if $M$ is connected, then a Killing field is determined by its $1$-jet at any point (this is actually an exercise in Moroianu's Lectures in Kähler Geometry).
My question is not about this. Flatness of $\overline{\nabla}$ would imply that given any $x \in M$ and $v \in T_pM$, there is a $\overline{\nabla}$-parallel section in a neighborhood of $x$ that realizes $v$. Motivated by that, I would like to compute the curvature $\overline{R}(X,Y)(Z,F)$. It has two components. The vector component vanishes (by a straightforward application of Bianchi's identity), while I get $$\begin{align*} & R(X,Y)F - \nabla_{X}R(Y,Z) + \nabla_{Y}R(X,Z) - R(X,\nabla_{Y}Z) \\ &\hspace{3cm} + R(Y,\nabla_{X}Z) + R(X,F(Y)) - R(Y,F(X)) + R([X,Y],Z) \end{align*}$$for the endomorphism component.
So:
Does the converse hold? Is it true that $\overline{\nabla}$-parallel sections define Killing fields? Or in other words, does $\nabla_X(\nabla\xi) = R(X,\xi)$ imply that $\xi$ is Killing?
Is there an efficient way to simplify the endomorphism component of $\overline{R}(X,Y)(Z,F)$? A lot of these terms seem to be somewhat distinct in nature. General references are also welcome.
