Exercise 5.6, do Carmo. Let $M$ be a Riemannian manifold of dimension two. Let $B_\delta(p)$ be a normal ball around the point $p \in M$ and consider the parametrized surface$$f(\rho, \theta) = \exp_p \rho v(\theta), \quad 0 < \rho < \delta, \quad -\pi < \theta < \pi,$$where $v(\theta)$ is a circle of radius $\delta$ in $T_pM$ parametrized by the central angle $\theta$.
(a) Show that $(\rho, \theta)$ are coordinates in an open set $U \subset M$ formed by the open ball $B_\delta(p)$ minus the ray $\exp_p(-\rho v(0))$, $0 < \rho < \delta$.
(b) Show that the coefficients $g_{ij}$ of the Riemannian metric in these coordinates are:$$g_{12} = 0, \quad g_{11} = \left|{{\partial f}\over{\partial \rho}}\right|^2 = |v(\theta)|^2 = 1, \quad g_{22} = \left|{{\partial f}\over{\partial \theta}}\right|^2.$$
(c) Show that, along the geodesic $f(\rho, 0)$, we have$$(\sqrt{g_{22}})_{\rho\rho} = -K(p)\rho + R(\rho), \text{ where }\lim_{\rho\to0}{{R(\rho)}\over\rho} = 0$$and $K(p)$ is the sectional curvature of $M$ at $p$.
(d) Prove that$$\lim_{\rho \to 0} {{(\sqrt{g_{22}})_{\rho\rho}}\over{\sqrt{g_{22}}}} = -K(p).$$
(a) As we are working in a normal ball about $p$, $\exp_p$ furnishes a diffeomorphism from a disc about the origin in $T_pM$ to $B_\delta(p)$. Moreover, the coordinates $(\rho,\theta)$ are given by the composition of the inverse of $\exp_p$ with $\psi_x^2+\psi_y^2$ and $\tan^{-1}\left( \psi_y/\psi_x \right)$, where $\psi_x$ and $\psi_y$ are the coordinates on $T_pM$. These are diffeomorphisms away from the ray $\exp_p(\rho v(0))$ for $0<\rho<\delta$.
(b) Let us compute the components $g_{ij}$ of the metric in polar
coordinates $(\rho,\theta)$. First, we have, by Gauss's lemma,
\begin{align*}
g_{11}&=g\left( \frac{\partial}{\partial \rho},\frac{\partial}{\partial\rho}\right)
= g\left( \frac{\partial f}{\partial \rho},\frac{\partial f}{\partial\rho} \right)
=\left|\frac{\partial f}{\partial\rho}\right|^2
\end{align*}
since, by definition, $\partial f/\partial\rho=f_*(\partial/\partial\rho)$. Moreover,
\begin{align*}
\left|(f_*)_{\rho v(\theta)}\left( \frac{\partial}{\partial\rho} \right)\right|^2&=\left|\frac{d}{dt}\bigg|_{t=0}\left( \exp_p\left( (\rho +t)v(\theta) \right) \right)\right|^2 =\left|\frac{d}{dt}\bigg|_{t=0}\gamma_p(\rho+t,v(\theta))\right|^2 \\ & =\left|\frac{d}{dt}\bigg|_{t=0}\gamma_p(t,v(\theta))\right|^2 =\left|v(\theta)\right|^2=1,
\end{align*}
where we have used the fact that $\gamma$ is a geodesic, and hence the
magnitude of its tangent vectors is constant. Next, again by Gauss's lemma,
we find that
\begin{align*}
g_{22} &= g\left( \frac{\partial}{\partial\theta}, \frac{\partial}{\partial\theta} \right)
=g\left( \frac{\partial f}{\partial\theta},\frac{\partial f}{\partial\theta} \right)
=\left|\frac{\partial f}{\partial\theta}\right|^2
\end{align*}
Finally, it follows from the proof of Gauss's lemma, c.f. page 70 of do Carmo, that
$$g_{12}=g_{21}=g\left( \frac{\partial}{\partial\rho},\frac{\partial}{\partial\theta} \right)=0.$$(c) Along the geodesic $f(\rho,0)$, it is clear that $\partial f/\partial\theta$ is a Jacobi field
and so we obtain via a Taylor expansion, see page 115 of do Carmo, that
$$\sqrt{g_{22}}=\left|\frac{\partial f}{\partial \theta}\right|=\rho-\frac{1}{6}K(p,\sigma)\rho^3+\tilde R(\rho),$$
where $\lim_{\rho\to 0}\tilde R(\rho)/\rho^3=0$. If we differentiate twice with respect
to $\rho$, we obtain
$$(\sqrt{g_{22}})_{\rho\rho}=-K(p,\sigma)\rho+R(\rho),$$
where $\lim_{\rho\to 0}R(\rho)/\rho=0$, as desired.
(d) Finally, note that
\begin{align*}
\lim_{\rho\to0}\frac{(\sqrt{g_{22}})_{\rho\rho}}{\sqrt{g_{22}}} = \lim_{\rho\to 0}\frac{-K(p,\sigma)\rho+R(\rho)}{\rho-K(p,\sigma)\rho^3/6+\tilde R(\rho)} =-K(p,\sigma).
\end{align*}
Exercise 5.7, do Carmo. Let $M$ be a Riemannian manifold of dimension two. Let $p \in M$ and let $V \subset T_pM$ be a neighborhood of the origin where $\exp_p$ is a diffeomorphism. Let $S_r(0) \subset V$ be a circle of radius $r$ centered at the origin, and let $L_r$ be the length of the curve $\exp_p(S_r)$ in $M$. Prove that the sectional curvature at $p \in M$ is given by$$K(p) = \lim_{r \to 0} {3\over\pi} {{2\pi r - L_r}\over{r^3}}.$$
Recall that the arc length of the curve $\text{exp}_p(S_r)$ is given by the integral$$L_r = \int_{-\pi}^\pi \sqrt{g\left({\partial\over{\partial\theta}}, {\partial\over{\partial\theta}}\right)}\,d\theta = \int_{-\pi}^\pi \sqrt{g_{22}}\,d\theta.$$The previous exercise showed that along $\text{exp}_p(rv(0))$, we have the identity$$\sqrt{g_{22}} = r - {1\over6}K(p, \sigma)r^3 + \tilde{R}(r).$$This in fact holds for all $\theta$, as $\sqrt{g_{22}}$ is independent of $\theta$. To see this, we can simply change coordinates to $(r, \theta - \theta_0)$, and the computation above yields the same expression for $\sqrt{g_{22}}(r, 0)$, which is now actually along $\text{exp}_p(rv(\theta_0))$. Hence, we obtain$$L_r = \int_{-\pi}^\pi \left(r - {1\over6}K(p, \sigma)r^3 + \tilde{R}(r)\right)d\theta = 2\pi r - {\pi\over3}K(p, \sigma)r^3 + 2\pi\tilde{R}(r).$$Dividing both sides by $r^3$ and taking the limit as $r \to 0$, we find that$$K(p, \sigma) = \lim_{r \to 0} {3\over\pi} {{2\pi r - L_r}\over{r^3}}.$$
First of all, I am sorry I could not find a complete reference for this fact: I think with some effort one can find something similar in Petersen, Riemannian geometry, Lee, Riemannian geometry, an introduction to curvature or Gallot, Hulin Lafontaine, Riemannian Geometry. But I could recover the result that follows ; there may be some typo. This is quite long but really is elementary.
Suppose $J$ is a Jacobi field along a geodesic $\gamma : I \to (M,g)$. We fixe some notations: we denote by $\sec$ the sectionnal curvature of $(M,g)$ and by $R_{\gamma}$ the tensor along $\gamma$ defined by $R_{\gamma}X = R(\gamma',X)\gamma'$. It is equivalent to say $g\left(R_{\gamma}X,X \right) = {\|X\|_g}^2\sec(\gamma',X)$. Recall the Jacobi equation is
$$
J'' +R_\gamma J = 0
$$
The function $t \in I \mapsto \|J(t)\|^2$ is smooth, and we have
\begin{align}
\dfrac{\mathrm{d}^2}{\mathrm{d}t^2}\left\|J\right\|^2 &= \dfrac{\mathrm{d}}{\mathrm{d}t} 2g\left(J',J\right)\\
&= 2g\left(J'',J \right) + 2g\left(J',J'\right) \\
&=-2g\left(R_\gamma J, J\right) + 2\|J'\|^2
\end{align}
Moreover, the left hand side can be computed another way for $t$ such that $J(t) \neq 0$:
\begin{align}
\dfrac{\mathrm{d}^2}{\mathrm{d}t^2}\left\|J\right\|^2 &= \dfrac{\mathrm{d}}{\mathrm{d}t} 2\|J\|\|J\|' \\
&= 2\left( {\|J\|'}^2 + \|J\|\|J\|'' \right)
\end{align}
and this tells us that for $t$ such that $J(t) \neq 0$, then
$$
2\left( {\|J\|'}^2 + \|J\|\|J\|'' \right) = -2g\left(R_\gamma J, J\right) + 2\|J'\|^2
$$
from which we deduce
$$
\left\|J \right\|'' = -\frac{g\left(R_{\gamma}J,J \right)}{\|J\|} = -\sec(\gamma',J)\|J\|
$$
Suppose $M$ has positive sectionnal curvature bounded from below along $\gamma$, say $\sec \geqslant \kappa^2 >0$, and define on $I$ $f(t) = \|J(0)\|\cos(\kappa t) + \|J(0)\|'\frac{\sin(\kappa t)}{\kappa}$. It is solution to the second order ODE $y'' = -\kappa^2 y$.
Now appears the trick: consider $g(t) = f(t) \|J(t)\|' - f'(t)\|J(t)\|$. Then on an interval containing $0$ on which $f(t) \geqslant 0$ and $J(t) \neq 0$ we have
\begin{align}
g' &= f\left(\|J\|'' + \kappa^2 \|J\| \right) \\
&= f \left(\|J\|'' + \sec(\gamma',J)\|J\| - \sec(\gamma',J)\|J\| +\kappa^2\|J\| \right) \\
&= f \left(\kappa^2 - \sec(\gamma',J) \right)\|J\| \geqslant 0
\end{align}
Hence, $g$ is non-decreasing and $g(0) = 0$. This shows that
$$
\forall t \text{ as above},~ g(t) \geqslant g(0) = 0
$$
which turns out to be
$$
\forall t \text{ as above},~ \frac{\|J\|'}{\|J\|} \leqslant \frac{f'}{f}
$$
integrating gives
$$
\ln \frac{\|J\|}{\|J(0)\|} \leqslant \ln \frac{f}{f(0)}
$$
and as $f(0) = \|J(0)\|$, we can deduce that for all $t$ such that all the above works
$$
0 \leqslant \|J(t)\| \leqslant f(t)
$$
Now, you can deduce vanishing properties of $J$ thanks to this inequality.
Also, the exact same study in case the sectionnal curvature is bounded from above $\sec \leqslant -\kappa^2 <0$ shows that
$$
\forall t,~ \|J(t)\| \geqslant \|J(0)\|\cosh (\kappa t) + \|J(0)\|' \frac{\sinh(\kappa t)}{\kappa}
$$
Comment: if $J(0) = 0$ one can adapt this proof to fix this. Also, $\|J\|$ may not be differentiable at $0$ if $J(0)=0$, but if the curvature is of constant sign, then above calculations imply that $\|J\|$ is concave or convex, thus differentiable from the right at zero and we are done.
Another comment: if $\kappa \geqslant 0$ without having a lower bound, we cannot say much, because the euclidean case shows that $J$ may not vanish for positive $t$.
Best Answer
Here's another possible way.
Let $v\in B_\epsilon(0)$ and $w\in T_v(T_pM)\cong T_pM$. Consider the variation of geodesics given by \begin{equation*} \gamma_s(t)=\exp_p\bigl(t(v+sw)\bigr) \end{equation*} It follows that \begin{equation*} J(t)=\frac{\partial\gamma_s}{\partial s}\biggr|_{s=0}=(d\exp_p)(tw) \end{equation*} is a Jacobi field along $\gamma_0$. Since sectional curvature is identically zero, $J$ satisfies the following PDE \begin{equation*} \frac{\partial^2J}{\partial t^2}=0 \end{equation*} Let $\{e_i\}$ be an orthnormal basis for $T_pM$ and extend it to a parallel frame along $\gamma_0$ such that $e_i(0)=e_i$. In terms of the frame, $J(t)=a^i(t)e_i(t)$ and $w=w^ie_i$. Solving the above equation, we have $J(t)=(w^it)e_i(t)$, so that \begin{equation*} w^ie_i(1)=(d\exp_p)(w) \end{equation*} Following the same argument, if $u\in T_v(T_pM)\cong T_pM$ and $u=u^ie_i$, then \begin{equation*} u^ie_i(1)=(d\exp_p)(u) \end{equation*} Therefore, \begin{align*} \langle(d\exp_p)(w),(d\exp_p)(u)\rangle &=w^iu^j\langle e_i(1),e_j(1)\rangle\\ &=w^iu^j\langle e_i,e_j\rangle\\ &=\langle w,u\rangle, \end{align*} proving that $\exp_p\colon B_\epsilon(0)\to B_\epsilon(p)$ is an isometry. Note that we used the fact that $\langle e_i(t),e_j(t)\rangle$ is indepdent of $t$.