About Sectional Curvature In a paper by Yann Ollivier:
Let $x$ be a point in $X$, $v$ a small tangent vector at $x$, $y$ the endpoint
of $v$, $w_x$ a small tangent vector at $x$, and $w_y$ the parallel transport of $w_x$ from $x$
to $y$ along $v$. If, instead of a Riemannian manifold, we were working in ordinary
Euclidean space, the endpoints $x_0$ and $y_0$ of $w_x$ and $w_y$ would constitute a rectangle
with $x$ and $y$. But in a manifold, generally these four points do not constitute a
rectangle any more.
Indeed, because of curvature, the two geodesics starting along $w_x$ and $w_y$ may
diverge from or converge towards each other. Thus, on a sphere (positive curvature),
two meridians starting at two points on the equator have parallel initial velocities,
yet they converge at the North (and South) pole. Since the initial velocities $w_x$ and
$w_y$ are parallel to each other, this effect is at second order in the distance along the
geodesics (Fig.).
Thus, let us consider the points lying at distance $\varepsilon $ from $x$ and $y$ on the geodesics
starting along $w_x$ and $w_y$, respectively. In a Euclidean setting, the distance between
those two points would be $|v|$, the same as the distance between $x$ and $y$. The
discrepancy from this Euclidean case is used as a definition of a curvature.

Definition(Sectional curvature). Let $(X, d)$ be a Riemannian manifold.
Let $v$ and $w_x$ be two unit-length tangent vectors at some point $x \in X$. Let $\varepsilon, \delta > 0$.
Let $y$ be the endpoint of $v$ and let $w_y$ be obtained by parallel transport of $w_x$ from $x$ to $y$. Then
$$ d(exp_{x} \varepsilon w_x, exp_{y} \varepsilon w_y) = \delta (1 −\frac{\varepsilon^2}{2} K(v,w)+ O(\varepsilon^3 +\delta \varepsilon^2))
 $$
when $\delta , \varepsilon \to 0$. This defines a quantity $K(v,w)$, which is the sectional curvature
at $x$ in the directions $(v,w)$.
Question1
How can I derive the formula $ d(exp_{x} \varepsilon w_x, exp_{y} \varepsilon w_y) = \delta (1 −\frac{\varepsilon^2}{2} K(v,w)+ O(\varepsilon^3 +\delta \varepsilon^2))
 $ from figure?
Question2
How this definition of sectional curvature can be derived from its usual definition ($K(v,w)=\frac{\langle R(v,w)w, v\rangle}{\langle v,v\rangle \langle w,w \rangle - \langle v,w \rangle ^2}$)
Thanks in advance for your time.
 A: The first equation doesn't really need to be 'derived' -- it is a definition.  The point $y$ is defined by travelling a small distance $\delta$ in the direction of $v$.  So
$$
\delta = d(x,y) = d\big(\exp_x(0),\exp_y(0)\big)~,
$$
and this gives you the term of zeroth order in $\epsilon$.  As explained in the text, there is no first-order term, and we then simply define $K(v,w)$ via the equation given.
As for the expression in terms of the Riemann tensor, your thinking is backwards.  You should start from the geometric definition of the sectional curvature, and derive the given relationship with the Riemann tensor.
A: Define a
variation $$ d(x,y)=\delta,\ c(t):=\exp_x \ tv,\ t\in [0,\delta ]
$$
$$ f(s,t):=\exp_{c(t)}\ sw(t),\ s\in [0,\varepsilon ] $$
where $w$ is a parallel vector field of $w$ along $c$ So $l(s):={\rm
length}\ f(s,\ ),\ l(0)=\delta$ so that $l(\varepsilon )=d(\exp_x
\varepsilon w_x,\exp_y \varepsilon w_y)$ Hence $$ l(s)
=\int_0^\delta |f_t| dt
$$ so that $$ l' =0,\ l'' = \int - K (c',w) $$
by variation formulas
That is, $$ l(s)=l(0) +\frac{s^2}{2} (\int -K) + O(s^3)$$
