A notion of nonpositive curvature for general metric spaces The proof of the following result should be done by using the second variation formula of geodesics but I do not know how to start or what is the main idea of the proof. (Lemma 3.7 in the paper: A notion of nonpositive curvature for general metric spaces)
Let $X,  Y\in T_mM$ be independent unit tangent  vectors at $m$. Then
$$\lim_{t\to 0}\frac{1}{t^2}\left(\frac{d(\exp_m(tX),\exp_m(tY))}{t\|X-Y\|}-1\right)
=-C(n,\angle(X,Y))K(X,Y),$$
where $K(X,Y)$ is the sectional curvature of Span$(X,Y)$ and $C(n,\angle(X,Y))$ is a constant  depending on the dimension of the manifold, $n$, and the angle between X and Y, $\angle(X,Y)$.
 A: (1) Consider variation $$ g(s,t),\ s\in [0,1],\ t\in [0,\varepsilon
)
$$ where $g(\ ,t)$ is a minimal geodesic from $$g(0,t)=\exp_m
((t_0-t)X) \ {\rm to}  \ g(1,t)=\exp_m((t_0-t)Y)$$
Hence we have energy
$$ E:= \frac{1}{2} \int_0^1 |g_s|^2\ ds = \frac{1}{2} L(t)^2,$$ $$ L(t):=d(\exp_m
(t_0-t)X,\exp_m (t_0-t)Y)$$
Then note that $g_t$ is a Jacobi field and we have $$ E' =
(g_s,g_t)|_0^1 $$
(2) Define $$ \theta_1:=\angle (g_s,g_t)(0,t),\ \theta_2:=\pi-\angle
(g_s,g_t)(1,t) $$
so that $$ E' =L'(0)L(0),\ L'(0)=(-\cos\ \theta_2-\cos\ \theta_1
)(0)
$$
$$ E'' = L(0)L''(0)  +
L'(0)^2
$$
$E'''=3L'L'' + LL'''$ so that $L(t)^2=  L^2 + 2LL' t + (LL'' +
(L')^2) t^2   + \frac{3L'L'' + LL''' }{3} t^3 +\cdots $
Hence $$ L(t)= L +L't + \frac{L''}{2}t^2 + \frac{L'''}{3} t^3+\cdots
$$
$$ \frac{L(t)}{t|X-Y| } = \frac{L(0) + L'(0)t}{t|X-Y| }  + \frac{L''
(0) }{2|X-Y|} t + \frac{L'''}{3|X-Y|} t^2 +  O(t^3) $$
That is $$ \frac{1}{t^2} \bigg( \frac{L(t)}{t|X-Y| } - \frac{L(0) +
L'(0)t}{t|X-Y| }  \bigg) =  \frac{L'' (0) }{2t|X-Y|} +CL''' + O(t)
$$
Here $L'''\rightarrow 0$ by considering Euclidean case.
By letting $t=\frac{1}{2}t_0$, $$ \frac{L(0) + L'(0) t}{
t|X-Y|}\rightarrow 1 $$ so that $$ \lim\ \frac{1}{t^2} \bigg(\frac{
L(t)}{t|X-Y|} - 1\bigg) =\lim\ C_2 \frac{L''(0)}{t} $$
(3) Note that $$ -\cos\ \theta_1= -\frac{1}{L}(g_t,g_s),\ -\cos\
 \theta_2= \frac{1}{L} (g_t,g_s) $$ so that $$
 \frac{L''}{t} = \frac{1}{t} (\frac{L'}{L^2} ( g_t,g_s)
 -\frac{1}{L} (g_t,\nabla_t g_s) - \frac{L'}{L^2} (g_t,g_s)(0,1) +
 \frac{1}{L}  (g_t,\nabla_t g_s)(0,1) ) $$
$$ = \frac{1}{t} \frac{L'}{L} \frac{
 (g_t,g_s)(0,0)-(g_t,g_s)(0,1)}{L}  - \frac{1}{t} \frac{(g_t,\nabla_t g_s)
 (0,0)-(g_t,\nabla_t g_s)(0,1)}{L} $$
$$ \sim -\frac{1}{t} \frac{L'}{L^2} (\nabla_s g_t,g_s) + \frac{1}{tL}
 ( R_{stst} + |\nabla_t g_s|^2 )
$$
Hence $$ \frac{L''}{t}\sim - \frac{L'}{t} \frac{L'}{L} + {\rm sec}\
(X,Y)C_3 + \frac{(L')^2}{tL} $$
