= Solution
For $v$ in a sufficiently small neighbourhood of $0\in T_pS$, let $\gamma_v$ be the geodesic satisfying $\gamma_v(0)=p$ and $\dot\gamma_v(0)=v$. The <exponential map> at $p$ is
$$
\exp_p(v)=\gamma_v(1).
$$
Choose an oriented orthonormal basis of $T_pS$, write $e(\theta)=(\cos\theta,\sin\theta)$, and define <geodesic polar coordinates> by
$$
\phi(r,\theta)=\exp_p\bigl(r e(\theta)\bigr).
$$
The <Gauss lemma> says that radial and angular coordinate curves are orthogonal. Since $r\mapsto\phi(r,\theta)$ is a unit-speed geodesic,
$$
|\phi_r|=1.
$$
Moreover, torsion-freeness of the surface connection gives $D_r\phi_\theta=D_\theta\phi_r$, and therefore
$$
\begin{aligned}
\frac{\partial}{\partial r}\langle\phi_r,\phi_\theta\rangle
&=\langle D_r\phi_r,\phi_\theta\rangle
+\langle\phi_r,D_r\phi_\theta\rangle\\
&=0+\langle\phi_r,D_\theta\phi_r\rangle
=\frac12\frac{\partial}{\partial\theta}|\phi_r|^2=0.
\end{aligned}
$$
At $r=0$, $\phi_\theta=0$, so $\langle\phi_r,\phi_\theta\rangle=0$ for every sufficiently small $r$. Thus the <first fundamental form> is
$$
\boxed{I=dr^2+G(r,\theta)\,d\theta^2},
\qquad G(r,\theta)=|\phi_\theta|^2,
$$
and $\sqrt G/r\to1$ as $r\downarrow0$ because $d(\exp_p)_0$ is the identity.
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