= Solution
Differentiate $f=g(S,S)$ twice, use <geodesic deviation>, and eliminate $g(DS,DS)$ with the conserved $K$:
$$
\ddot f=2g(DS,DS)+2g(S,D^2S)=2K+4\kappa f,\qquad \boxed{\ddot f-\frac R3f=2K.}
$$
The most informative way to interpret this <linear differential equation> is to express $S$ in a <parallel-propagated orthonormal frame> along the <timelike geodesic>. Its three spatial components satisfy $\ddot S^i=\kappa S^i$.
If $R>0$, put $H=\sqrt{R/12}$. With constant spatial vectors $A,B$ in that frame,
$$
S=Ae^{H\tau}+Be^{-H\tau},\qquad f=|A|^2e^{2H\tau}+2A\cdot B+|B|^2e^{-2H\tau},\qquad K=-4H^2A\cdot B.
$$
Thus \b[generic deviations grow in length like $e^{H\tau}$], and their squared separation grows like $e^{2H\tau}$. Initially comoving neighbors, $DS(0)=0$, have $S(\tau)=S(0)\cosh(H\tau)$ and exhibit this growth. The unqualified printed assertion is too strong: the nonzero <Jacobi field> $S=Be^{-H\tau}$ has $K=0$ and shrinks exponentially. Such a <Jacobi field> is realizable by varying initial position and initial unit timelike velocity, with initial relative velocity $DS(0)=-HB$.
If $R<0$, put $\omega=\sqrt{-R/12}$. Then
$$
S=A\cos(\omega\tau)+B\sin(\omega\tau),\qquad \ddot f+4\omega^2f=2K.
$$
\b[The deviations are bounded and oscillatory, rather than exponentially growing.] The squared separation is periodic with period $\pi/\omega$ or is constant. Initially comoving neighbors have $S(\tau)=S(0)\cos(\omega\tau)$ and focus at $\tau=\pi/(2\omega)$; arbitrary vector initial data need not focus together. For completeness, $R=0$ gives $S=A+B\tau$ and $f=|A+B\tau|^2$. These conclusions concern infinitesimal <geodesic deviation> on the domain of the variation, not arbitrarily large finite separations.
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