Does the value of the covariant derivative at a point of the metric tensor depend only on the involved tangent vectors? 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.
 A: After having thought a bit about the above problem, and thanks to the remarks of Zhen Lin, I think I am in a position to answer my question in the affirmative.
It is true that, whenever you have an $r$-multilinear map 
$$A: \underbrace{\Gamma(TM) \times ... \times \Gamma(TM)}_{r-times} \longrightarrow \Gamma(TM)$$ such that for all $f \in C^{\infty}(M)$ we also have $$A(X_{1},...,X_{i-1},fX_{i},X_{i+1},...,X_{r})= fA(X_{1},...,X_{r})$$
in every component of $A$ then there exists an $(r,1)$-tensor field $B$ such that $A(X_{1},...,X_{r})(p)=B_{p}(X_{1}(p),...,X_{r}(p))$. In particular the value of $A(X_{1},...,X_{r})(p)$ does depend only on the tangent vectors $X_{1}(p),...,X_{r}(p)$.
The same, however is true, when we consider a (smooth) curve $c:[0,1] \rightarrow M$ and replace $\Gamma(TM)$ by $\Gamma_{c}(TM)$ ($=$vector fields along $c$) in the above reasoning. One then simply has to check that
$$\frac{d}{dt}|_{t=0} \; g(fY,Z)-g \left( \frac{\nabla}{dt}|_{t=0} \; fY,Z \right)-g \left( fY, \frac{\nabla}{dt}|_{t=0} \; Z \right)= f \left( \frac{d}{dt}|_{t=0} \; g(Y,Z)-g \left( \frac{\nabla}{dt}|_{t=0} \; Y,Z \right)-g \left( Y, \frac{\nabla}{dt}|_{t=0} \; Z \right) \right)$$
and
$$
\frac{d}{dt}|_{t=0} \; g(Y,fZ)-g \left( \frac{\nabla}{dt}|_{t=0} \; Y,fZ \right)-g \left( Y, \frac{\nabla}{dt}|_{t=0} \; fZ \right)= f \left( \frac{d}{dt}|_{t=0} \; g(Y,Z)-g \left( \frac{\nabla}{dt}|_{t=0} \; Y,Z \right)-g \left( Y, \frac{\nabla}{dt}|_{t=0} \; Z \right) \right)
$$
for all $f \in C^{\infty}([0,1])$, but this is easy.
Hence indeed $$\frac{d}{dt}|_{t=0} \; g(Y,Z)-g \left( \frac{\nabla}{dt}|_{t=0} \; Y,Z \right)-g \left( Y, \frac{\nabla}{dt}|_{t=0} \; Z \right)$$ does depend only on $Y(0)$ and $Z(0)$.
