Derivative of a parallel translation inside a metric

Let $M$ be a riemannian manifold with metric $g$ and a connection $\nabla$ on $M$. Let $X,Y$ two vector fields along a curve $\gamma$ on $M$. Let $$\tau_{t,s}:T_{\gamma(s)}M\to T_{\gamma(t)}M$$ the parallel translation along $\gamma$. We consider the fonction $$F(t)=g(\gamma(t))(\tau_{t,s} X(\gamma(s)), \tau_{t,s} Y(\gamma(s))).$$ Then $$F'(s)=g(\gamma(t))\left(\frac{d}{dt}\bigg|_{t=s}\tau_{t,s} X(\gamma(s)), \tau_{t,s}Y(\gamma(s))\right)+g(\gamma(t))\left(\frac{d}{dt}\bigg|_{t=s}\tau_{t,s} Y(\gamma(s)), \tau_{t,s}X(\gamma(s))\right)+\dot\gamma\left[g(X,Y)\right](\gamma(s)) \ (\star).$$

I do not understand why this identity holds. Here is what I have tried so far:

Let $$a(t)=\tau_{t,s} X(\gamma(s))=a_i(t)\frac{\partial}{\partial x_i}\bigg|_{\gamma(t)}\text{ and }b(t)=\tau_{t,s} Y(\gamma(s))=b_j(t)\frac{\partial}{\partial x_i}\bigg|_{\gamma(t)},$$ so $F(t)=g(\gamma(t))(a(t),b(t))$. If we let $$g_{ij}(t)=g(\gamma(t))\left(\frac{\partial}{\partial x_i}\bigg|_{\gamma(t)}, \frac{\partial}{\partial x_j}\bigg|_{\gamma(t)}\right),$$ we find that $F(t)=a_i(t)b_j(t)g_{ij}(t)$, thus the derivative of $F$ is $$F'(s)=a_i'(s)b_j(s)g_{ij}(s)+a_i(s)b_j'(s)g_{ij}(s)+a_i(s)b_j(s)g'_{ij}(s).$$ The first two terms give the first two terms of $(\star)$, but what about the last one ? We have that $$\dot\gamma\left[g(X,Y)\right](\gamma(s))=\frac{d}{dt}\bigg|_{t=s} g(\gamma(t))(X(\gamma(t)),Y(\gamma(t))),$$ right ? This does not seem equal to $a_i(s)b_j(s)g'_{ij}(s)$... ?

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Your definition of $F(t)$ doesn't make sense unless $s = \epsilon$. This in turn causes problems with your expression for $F'(s)$. – treble Dec 23 '12 at 22:18
You mean $t$ close enough to $s$ ? It was implicitly assumed. – Klaus Dec 24 '12 at 10:02
My fault, I thought your subscript $s$ was an $\epsilon$ (hard for me to see those tiny subscripts). It makes sense now. – treble Dec 24 '12 at 19:14