Suppose $X,Y$ are vector fields along a curve $c:s\mapsto c(s)$, i.e. $X(s)\in T_{c(s)}M$ and $s\mapsto X(s)$ is smooth. Show that
$$
\ttd s\la X(s),Y(s)\ra
=\la\bnabla X(s),Y(s)\ra+\la X(s),\bnabla Y(s)\ra~.
$$
2. If $f$ is a smooth functions of $s$, then
$$
\bnabla(fX)(s)=f^\prime(s)X(s)+f(s)\bnabla X(s)~.
$$
3. For any $v\in T_{c(0)}M$ the mapping $J:s\mapsto P_s(v)$ is a vector field along $c$ and $\bnabla J(s)=0$. Conversely if $X$ is a vector field along $c$ such that $\bnabla X(s)=0$, then $X(s)=P_s(X(0))$. Thus the parallel transport of a vector $v\in T_{c(0)}M$ along $c$ is uniquely defined by $\bnabla X(s)=0$ and $X(0)=v$.
1. For simplicity we compute the derivative at $s=0$. So let $P_s$ denote parallel transport along $c$, then
\begin{eqnarray*}
\la X(s),Y(s)\ra-\la X(0),Y(0)\ra
&=&\la X(s)-P_sX(0)+P_sX(0),Y(s)-P_sY(0)+P_sY(0)\ra-\la X(0),Y(0)\ra\\
&=&\la X(s)-P_sX(0),Y(s)-P_sY(0)\ra
+\la P_sX(0),Y(s)-P_sY(0)\ra\\
&&+\la X(s)-P_sX(0),P_sY(0)\ra
+\la P_sX(0),P_sY(0)\ra
-\la X(0),Y(0)\ra\\
\end{eqnarray*}
Since $P_s$ is an isometry it follows that $\la P_sX(0),P_sY(0)\ra=\la X(0),Y(0)\ra$. Finally, as $s\mapsto P_s$ is continuous and $P_0=1$ we conclude that
$$
\bnabla X(0)
=\lim_{s\to0}\frac1s(P_s^{-1}X(s)-X(0))
=\lim_{s\to0}P_s^{-1}\Big(\frac1s(X(s)-P_sX(0))\Big)
=\lim_{s\to0}\frac1s(X(s)-P_sX(0))~.
$$
Hence we get by dividing by $s$ in the limit $s\to0$:
$$
\ttdl s0\la X(s),Y(s)\ra
=\la\bnabla X(0),Y(0)\ra+\la X(0),\bnabla Y(0)\ra~.
$$
2. For any vector field $Y$ along $c$ we have by 1.:
$$
\la\bnabla(fX),Y\ra+\la fX,\bnabla Y\ra
=\ttd s\la(fX),Y\ra
=f^\prime\la X,Y\ra+f\ttd s\la X,Y\ra
=\la f^\prime X+f\bnabla X,Y\ra+\la fX,\bnabla Y\ra
$$
i.e. $\la\bnabla(fX),Y\ra=\la f^\prime X+f\bnabla X,Y\ra$ and as $Y$ was arbitrary: $\bnabla(fX)=f^\prime X+f\bnabla X$.
3. Suppose $X$ is a vector field along $c$ such that $\bnabla X(s)=0$, then for the vector fields $Y(s)\colon=P_s(v)$ and $J(s)\colon=P_s(X(0))$ we get
$$
\ttd s\la X(s)-J(s),Y(s)\ra=0
$$
Substituting an orthonormal basis $e_0,\ldots,e_n$ for $v$ we conclude that $X(s)-J(s)=0$, because $P_s(e_0),\ldots,P_s(e_n)$ is an orthonormal basis for $(T_{c(s)}M,\la.,.\ra_{c(s)})$.