[proofplan]
The proof reduces the equality case to a scalar comparison for the length $f(t)=|J(t)|_g$. Since there are no conjugate points, $J$ does not vanish on $(0,t_0]$, so we may write $J=fE$ with $E$ a unit field perpendicular to $\dot{\gamma}$. The Jacobi equation gives a differential inequality for $f$ against the model solution $s=\operatorname{sn}_k$, and the Wronskian of $f$ and $s$ is monotone. Equality at $t_0$ forces the monotone ratio $f/s$ to be identically equal to $1$, and then the nonnegative error terms in the scalar equation vanish, giving both parallelness of $E$ and equality of the radial sectional curvature.
[/proofplan]
[step:Write the Jacobi field as its length times a unit perpendicular field]
Let
\begin{align*}
s:[0,\infty)&\to\mathbb{R}\\
t&\mapsto \operatorname{sn}_k(t)
\end{align*}
denote the model solution in the statement. Since $D_tJ(0)=E_0$ and $|E_0|_g=1$, the [Jacobi field](/page/Jacobi%20Field) $J$ is not identically zero. Because $\gamma|_{[0,R]}$ has no [conjugate point](/page/Conjugate%20Point) to $\gamma(0)$, the nonzero Jacobi field $J$ cannot vanish at any $t\in(0,R]$. Hence the function
\begin{align*}
f:(0,t_0]&\to(0,\infty)\\
t&\mapsto |J(t)|_g
\end{align*}
is well-defined and positive.
Define
\begin{align*}
E:(0,t_0]&\to TM\\
t&\mapsto \frac{J(t)}{f(t)}\in T_{\gamma(t)}M.
\end{align*}
Then $E$ is a smooth unit vector field along $\gamma|_{(0,t_0]}$, and since $J(t)\perp \dot{\gamma}(t)$, we have $E(t)\perp \dot{\gamma}(t)$ for every $t\in(0,t_0]$. Thus
\begin{align*}
J(t)=f(t)E(t)
\end{align*}
for every $t\in(0,t_0]$.
[/step]
[step:Derive the scalar Jacobi inequality for the length]
For $t\in(0,t_0]$, define the radial sectional curvature along the plane spanned by $\dot{\gamma}(t)$ and $E(t)$ by
\begin{align*}
K(t):=\operatorname{sec}_g(\dot{\gamma}(t),E(t)).
\end{align*}
Let $R$ denote the Riemann curvature tensor of $(M,g)$, viewed as the map
\begin{align*}
R:TM\times TM\times TM&\to TM\\
(X,Y,Z)&\mapsto R(X,Y)Z.
\end{align*}
Since $E(t)$ has unit length and is perpendicular to the unit vector $\dot{\gamma}(t)$, this is equivalently
\begin{align*}
K(t)=g(R(E(t),\dot{\gamma}(t))\dot{\gamma}(t),E(t)),
\end{align*}
with the curvature sign convention determined by the Jacobi equation
\begin{align*}
D_t^2J+R(J,\dot{\gamma})\dot{\gamma}=0.
\end{align*}
Because $g(E,E)=1$, differentiating gives $g(D_tE,E)=0$. Differentiating once more gives
\begin{align*}
g(D_t^2E,E)=-|D_tE|_g^2.
\end{align*}
Using $J=fE$, we compute
\begin{align*}
D_t^2J=f''E+2f'D_tE+fD_t^2E.
\end{align*}
Taking the $g$-[inner product](/page/Inner%20Product) with $E$ and using the Jacobi equation gives
\begin{align*}
0
&=g(D_t^2J+R(J,\dot{\gamma})\dot{\gamma},E)\\
&=f''+2f'g(D_tE,E)+f\,g(D_t^2E,E)+f\,g(R(E,\dot{\gamma})\dot{\gamma},E)\\
&=f''-f|D_tE|_g^2+fK.
\end{align*}
Therefore
\begin{align*}
f''+k f=f\bigl(|D_tE|_g^2+k-K\bigr).
\end{align*}
The curvature hypothesis gives $K(t)\leq k$, so the function
\begin{align*}
q:(0,t_0]&\to[0,\infty)\\
t&\mapsto |D_tE(t)|_g^2+k-K(t)
\end{align*}
is nonnegative and satisfies
\begin{align*}
f''+k f=qf.
\end{align*}
[guided]
The goal is to isolate exactly where equality can fail in Rauch comparison. Since $J$ never vanishes on $(0,t_0]$, we can separate its size from its direction:
\begin{align*}
J(t)=f(t)E(t),\qquad f(t)=|J(t)|_g,\qquad |E(t)|_g=1.
\end{align*}
The field $E$ records the direction of $J$, while $f$ records its length.
We now compute the scalar equation satisfied by $f$. First, $g(E,E)=1$, so differentiating along $\gamma$ gives
\begin{align*}
0=\frac{d}{dt}g(E,E)=2g(D_tE,E),
\end{align*}
hence $g(D_tE,E)=0$. Differentiating this identity once more gives
\begin{align*}
0=\frac{d}{dt}g(D_tE,E)=g(D_t^2E,E)+|D_tE|_g^2,
\end{align*}
so
\begin{align*}
g(D_t^2E,E)=-|D_tE|_g^2.
\end{align*}
Using $J=fE$, the covariant product rule gives
\begin{align*}
D_t^2J=f''E+2f'D_tE+fD_t^2E.
\end{align*}
The Jacobi equation is
\begin{align*}
D_t^2J+R(J,\dot{\gamma})\dot{\gamma}=0.
\end{align*}
Taking the inner product with $E$ gives
\begin{align*}
0
&=g(D_t^2J+R(J,\dot{\gamma})\dot{\gamma},E)\\
&=f''+2f'g(D_tE,E)+f\,g(D_t^2E,E)+f\,g(R(E,\dot{\gamma})\dot{\gamma},E)\\
&=f''-f|D_tE|_g^2+fK(t),
\end{align*}
where
\begin{align*}
K(t):=\operatorname{sec}_g(\dot{\gamma}(t),E(t)).
\end{align*}
Thus
\begin{align*}
f''+k f=f\bigl(|D_tE|_g^2+k-K(t)\bigr).
\end{align*}
This identity displays the two possible sources of strict inequality: the direction $E$ may rotate, measured by $|D_tE|_g^2$, and the radial sectional curvature may be strictly smaller than $k$, measured by $k-K(t)$. Both terms are nonnegative because the curvature hypothesis says $K(t)\leq k$.
[/guided]
[/step]
[step:Use the Wronskian to prove monotonicity of the quotient $f/s$]
Define the Wronskian-type function
\begin{align*}
W:(0,t_0]&\to\mathbb{R}\\
t&\mapsto f'(t)s(t)-f(t)s'(t).
\end{align*}
Since $s''+ks=0$ and $f''+kf=qf$, we have
\begin{align*}
W'(t)
&=f''(t)s(t)+f'(t)s'(t)-f'(t)s'(t)-f(t)s''(t)\\
&=f''(t)s(t)-f(t)s''(t)\\
&=\bigl(q(t)f(t)-kf(t)\bigr)s(t)+kf(t)s(t)\\
&=q(t)f(t)s(t).
\end{align*}
For $t\in(0,t_0]$, the functions $q$, $f$, and $s$ are nonnegative, with $f(t)>0$ and $s(t)>0$. Hence
\begin{align*}
W'(t)\geq 0.
\end{align*}
Because $J(0)=0$ and $D_tJ(0)=E_0$, Taylor expansion along $\gamma$ gives functions $a,b:(0,t_0]\to\mathbb{R}$ satisfying
\begin{align*}
\lim_{t\downarrow0}a(t)=0,\qquad \lim_{t\downarrow0}b(t)=0,
\end{align*}
such that
\begin{align*}
f(t)=t\bigl(1+a(t)\bigr),\qquad f'(t)=1+b(t)
\end{align*}
as $t\downarrow0$. Since $s(0)=0$ and $s'(0)=1$, Taylor expansion for the model solution gives a function $c:(0,t_0]\to\mathbb{R}$ satisfying
\begin{align*}
\lim_{t\downarrow0}c(t)=0
\end{align*}
such that
\begin{align*}
s(t)=t\bigl(1+c(t)\bigr),\qquad s'(t)\to1
\end{align*}
as $t\downarrow0$. Therefore
\begin{align*}
\lim_{t\downarrow 0}\frac{f(t)}{s(t)}=1.
\end{align*}
Moreover,
\begin{align*}
\lim_{t\downarrow0}W(t)
&=\lim_{t\downarrow0}\bigl(f'(t)s(t)-f(t)s'(t)\bigr)\\
&=0.
\end{align*}
For $t\in(0,t_0]$,
\begin{align*}
\frac{d}{dt}\left(\frac{f(t)}{s(t)}\right)
=\frac{f'(t)s(t)-f(t)s'(t)}{s(t)^2}
=\frac{W(t)}{s(t)^2}.
\end{align*}
Since $W$ is nondecreasing and has limit $0$ as $t\downarrow0$, we have $W(t)\geq0$. Thus $f/s$ is nondecreasing on $(0,t_0]$, and
\begin{align*}
\frac{f(t)}{s(t)}\geq1
\end{align*}
for every $t\in(0,t_0]$.
[/step]
[step:Force every nonnegative error term to vanish from equality at $t_0$]
The hypothesis $|J(t_0)|_g=s(t_0)$ says
\begin{align*}
\frac{f(t_0)}{s(t_0)}=1.
\end{align*}
Since $f/s$ is nondecreasing on $(0,t_0]$ and has limit $1$ as $t\downarrow0$, it follows that
\begin{align*}
\frac{f(t)}{s(t)}=1
\end{align*}
for every $t\in(0,t_0]$. Hence
\begin{align*}
f(t)=s(t)
\end{align*}
for every $t\in(0,t_0]$.
The derivative of $f/s$ is identically zero, so $W(t)=0$ for every $t\in(0,t_0]$. Since
\begin{align*}
W'(t)=q(t)f(t)s(t)
\end{align*}
and $f(t)>0$, $s(t)>0$ on $(0,t_0]$, we obtain
\begin{align*}
q(t)=0
\end{align*}
for every $t\in(0,t_0]$. By the definition of $q$,
\begin{align*}
|D_tE(t)|_g^2+k-K(t)=0.
\end{align*}
Both summands are nonnegative, so
\begin{align*}
|D_tE(t)|_g^2=0,\qquad K(t)=k
\end{align*}
for every $t\in(0,t_0]$. Therefore $D_tE(t)=0$ on $(0,t_0]$, and
\begin{align*}
\operatorname{sec}_g(\dot{\gamma}(t),E(t))=k
\end{align*}
for every $t\in(0,t_0]$.
[/step]
[step:Extend the parallel direction to the initial point and identify $J$]
Since $D_tE=0$ on $(0,t_0]$, the field $E$ is parallel along $\gamma|_{(0,t_0]}$. Parallel transport preserves length and orthogonality to $\dot{\gamma}$, because $D_t\dot{\gamma}=0$ and $D_tE=0$.
We have already shown $J(t)=s(t)E(t)$ for $t\in(0,t_0]$. Since $s'(0)=1$ and $D_tJ(0)=E_0$, taking the limit as $t\downarrow0$ gives
\begin{align*}
\lim_{t\downarrow0}E(t)
=\lim_{t\downarrow0}\frac{J(t)}{s(t)}
=E_0.
\end{align*}
Define $E(0):=E_0$. Then $E$ is the parallel extension of $E_0$ along $\gamma|_{[0,t_0]}$, has unit length, remains perpendicular to $\dot{\gamma}$, and satisfies
\begin{align*}
J(t)=s(t)E(t)=\operatorname{sn}_k(t)E(t)
\end{align*}
for every $t\in[0,t_0]$. The curvature equality proved above gives
\begin{align*}
\operatorname{sec}_g(\dot{\gamma}(t),E(t))=k
\end{align*}
for every $t\in(0,t_0]$. This is precisely the asserted equality case.
[/step]
