= Solution
For an affinely parametrized <geodesic> $\gamma$, put $T=\dot\gamma$ and $D_t=\nabla_T$. A <Jacobi field> is a smooth vector field $J$ along $\gamma$ satisfying
$$
D_t^2J+R(J,T)T=0.
$$
This is a linear second-order equation, so $J$ is uniquely determined by $J(0)$ and $D_tJ(0)$. The defining sign uses the curvature convention in the preceding solution.
To prove the <realization of Jacobi fields by geodesic variations>, write $p=\gamma(0)$, $V=T(0)$, $A=J(0)$ and $B=D_tJ(0)$. Choose a small smooth curve $p(s)$ with $p(0)=p$ and $p'(0)=A$, for instance $p(s)=\exp_p(sA)$. Let $P_s:T_pM\to T_{p(s)}M$ be <parallel transport> along this curve and set $V(s)=P_s(V+sB)$. Then $V(0)=V$ and $\nabla_sV(s)|_{s=0}=B$. Define
$$
f(t,s)=\exp_{p(s)}\bigl(tV(s)\bigr).
$$
For sufficiently small $\varepsilon>0$, this is defined and smooth for every $(t,s)\in[0,1]\times(-\varepsilon,\varepsilon)$. No completeness assumption is needed: smooth dependence for the <geodesic> ordinary differential equation gives an open set of initial data whose solutions exist throughout the compact interval $[0,1]$ around the given solution. The given <geodesic> also extends slightly beyond its endpoints by local existence.
Every $t$-curve of $f$ is a <geodesic> and $f(t,0)=\gamma(t)$. If $\widehat J=\partial_sf|_{s=0}$, torsion-freeness gives $\nabla_s\partial_tf=\nabla_t\partial_sf$. Since $\nabla_t\partial_tf=0$, commuting <covariant derivatives> yields
$$
0=\nabla_s\nabla_t\partial_tf
=\nabla_t^2\partial_sf+R(\partial_sf,\partial_tf)\partial_tf.
$$
Thus $\widehat J$ satisfies the Jacobi equation. Moreover $\widehat J(0)=A$ and $D_t\widehat J(0)=B$. Uniqueness of the linear equation proves \b[$\partial_sf(t,0)=J(t)$].
If $J(0)=0$, keep the initial point fixed and choose initial velocities $V+sB$. Differentiating $f(t,s)=\exp_p(tV+stB)$ gives the general formula for <Jacobi fields vanishing at their initial point>:
$$
\boxed{J(t)=(d\exp_p)_{tV}(tB),\qquad B=D_tJ(0)\in T_pM}.
$$
Here $T_{tV}(T_pM)$ is identified with $T_pM$. Every choice of $B$ gives a <Jacobi field> with zero initial value, and uniqueness shows that these are all such fields. The factor $t$ is part of the formula.
Points $\gamma(0)$ and $\gamma(L)$, with $L>0$, are conjugate along this <geodesic> if there is a nonzero <Jacobi field> with $J(0)=J(L)=0$. Equivalently, $(d\exp_p)_{LV}$ has a nontrivial kernel: for a field with zero initial value, $J$ is nonzero exactly when $B\ne0$. This definition refers to the particular <geodesic> segment, not merely to the two endpoints.
Now suppose the sectional curvatures along the segment are nonpositive. Let $h(t)=|J(t)|^2$. <Metric compatibility> and the Jacobi equation give
$$
h''(t)=2|D_tJ|^2-2g(R(J,T)T,J).
$$
When $J,T$ are independent, the curvature term is $K(J,T)(|J|^2|T|^2-g(J,T)^2)\le0$; it is zero when they are dependent. Therefore $h''\ge0$. If $J$ vanishes at both endpoints, convexity gives $h(t)\le(1-t/L)h(0)+(t/L)h(L)=0$, whereas $h(t)\ge0$ by definition. Hence $h=0$ and $J=0$. \b[There are no conjugate points along the segment.] This proves that <nonpositive sectional curvature excludes conjugate points>; it does not require the manifold to be complete or simply connected.
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