Picture below is from do Carmo's Riemannian Geometry, I don't know how to show the red line. The $S$ is shape operator, assuming local extension of $\gamma'(0)$ is $N$ which normal to $M$, then $$ S_{\gamma'(0)} J(0) = -(\nabla_{J(0)}N)^T $$ where $\nabla$ is connection of $M$, and $T$ means the tangential component.
What I get from Deane: Since his mark is a little confused, I'll explain the mark first. Let $\gamma(t)$ is the geodesic in picture below. $f(s,t)$ is the extand of $\gamma(t)$ such that $$ f(0,t) =\gamma(t),~~~ \partial_t f(0,0) =\gamma'(0),~~~ \partial_t f(s,0) \bot\sum\nolimits_\epsilon $$ Then, $J(t)=\partial_s f(0,t)$ is Jacobi field along $\gamma(t)$, and to be less precise, there is $$ S_{\gamma'(0)} J(0) = -(\nabla_{J(0)}\partial_t f(s,0))^T $$ For showing $S_{\gamma'(0)} J(0)=0$, $\forall v\in T_{\gamma(0)}\sum\nolimits_\epsilon$, there is $$ \langle S_{\gamma'(0)} J(0), v\rangle = \langle -(\nabla_{J(0)}\partial_t f(s,0))^T, v \rangle = \langle -\nabla_{J(0)}\partial_t f(s,0), v \rangle = \langle \partial_t f(s,0), \nabla_{J(0)} v \rangle $$ Then, I feel there is $\nabla_{J(0)} v\in T_{\gamma(0)}\sum\nolimits_\epsilon$. If so, there is $\langle S_{\gamma'(0)} J(0), v\rangle = 0$. But I can't explain why $\nabla_{J(0)} v\in T_{\gamma(0)}\sum\nolimits_\epsilon$.
Here's the story. I'm pretty sure it's all in the book somewhere.
First, let $\Sigma \subset M$ be any hypersurface. For each $x \in \Sigma$, let $\nu(x)$ be a unit normal to $\Sigma$ such that $\nu$ always points to the same side of $\Sigma$. This, of course, is the Gauss map. Recall that given any $X \in T_x\Sigma$, the shape operator is defined to be $$ S(X) = \nabla_X\nu. $$
You can then define the exponential map from $\Sigma$ (rather than from a point) to be the map $$ E: \Sigma\times(-\delta,\delta) \rightarrow M, $$ where for each $x \in \Sigma$, the curve $$ \gamma_x(t) = E(x,t) $$ is a unit speed geodesic. Given $x \in \Sigma$ and $X \in T_x\Sigma$, let $c: (-\epsilon,\epsilon) \rightarrow \Sigma$ be a curve such that $c(0)=x$ and $c'(0) = X$. Consider the $1$-parameter family of geodescs \begin{align*} C: (-\epsilon,\epsilon)\times(-\delta,\delta) &\rightarrow M\\ (s,t) &\mapsto E(c(s),t). \end{align*} Then $$ J(t) = \left.\partial_sC(s,t)\right|_{s=0} $$ is a Jacobi field along the geodesic $\gamma_x$ such that $$ J(0) = c'(0) = X $$ and \begin{align*} J'(0) &= \nabla_{\partial_t C}\partial_sC(0,0)\\ &= \nabla_{\partial_sC(0,0)}\partial_tC(0,0)\\ &= \nabla_X\nu\\ &= S(X). \end{align*}
Now let $$ T = \{ v\in T_xM\ :\ \langle v,\gamma_x'(0)\rangle = 0 \}$$ and let $\Sigma$ be the hypersurface parameterized by the map \begin{align*} T &\rightarrow M\\ v &\mapsto \exp_x(v). \end{align*} In particular, for each $X \in T_x\Sigma$, the constant speed geodesic $$ c(s) = \exp_x sX $$ is in $\Sigma$. Therefore, \begin{align*} \langle X, S(X)\rangle &= \langle c'(0), \nabla_{c'(0)}\nu\rangle\\ &= \left.\partial_s\right|_{s=0}\langle c'(s),\nu(c(s))\rangle - \langle \nabla_{c'}c',\nu\rangle\\ &= 0. \end{align*} Since this holds for all $X \in T_x\Sigma$, it follows that $$ S(v) = 0 $$ for all $v \in T_x\Sigma$.