Let $\nabla$ be an affine connection on a pseudo-Riemannian manifold $(M,g)$. Let $c:[0,1] \rightarrow M$ be a differentiable curve and consider vector fields $Y,Z$ along $c$. Is it true that the expression
$\frac{d}{dt}|_{t=0} \; g(Y,Z)-g \left( \frac{\nabla}{dt}|_{t=0} \; Y,Z \right)-g \left( Y, \frac{\nabla}{dt}|_{t=0} \; Z \right)$
does depend only on the tangent vectors $Y(0)$ and $Z(0)$?
My motivation for this claim comes from the fact that, for all vector fields $X,Y,Z \in \Gamma(TM)$ and all points $p \in M$, the covariant derivative of the metric tensor $(\nabla g)(X,Y,Z)(p):= X_{p}g(Y,Z)-g \left( \nabla_{X_{p}}Y,Z_{p} \right)-g \left( Y_{p}, \nabla_{X_{p}}Z \right)$ depends only on the tangent vectors $X_{p},Y_{p},Z_{p}$. If one chooses a curve $\tilde{c}:[0,1] \rightarrow M$ such that $\tilde{c}(0)=p$, $\tilde{c}'(0)=X_{p}$ one is led to the first expression above, so it seems that my question can be answered in the affirmative, if the vector fields $Y,Z$ along $\tilde{c}$ are of the form $\tilde{Y} \circ \tilde{c}$, $\tilde{Z} \circ {c}$ with $\tilde{Y}, \tilde{Z} \in \Gamma(TM)$.
But what about the general case?
I might add that the original exercise I wanted to solve was to prove that the metric tensor is parallel (i.e. $\nabla g =0$), if and only if for every curve $c$ every parallel transport along $c$ is an isometry. However, I came across my question above while attempting to prove this and found the question interesting in its own right.