I believe that the focal locus is the same thing as the conjugate locus in Riemannian geometry. Given a codimension 1 submanifold $S \subset \mathbb{R}^n$, the exponential map $S \times \mathbb{R} \rightarrow \mathbb{R}^n$ is given by $(x, t) \mapsto x + t\gamma(x)$, where $\gamma$ is the Gauss map. The cut locus is the closure of all points in $\mathbb{R}^n$ that have more than one pre-image. The focal points is the closure of all points where the map is not a diffeomorphism.
Given any point where the curvature $\kappa$ is nonzero on a smooth curve in the plane, there is a corresponding point on the focal locus at distance $1/|\kappa|$ on the side of the curve that is inwardly curved.
I believe that the focal locus is always contained in the cut locus.
ADDED: (Corrected definition of map above)...If you differentiate the exponential map (the one I define above), then since the differential of the Gauss map is the second fundamental form, call it $A$, then you see that if a point $y$ lies in the focal locus, there exists $x \in S$, $t \in \mathbb{R}$, and a nonzero $v \in \mathbb{R}^n$ such that $v + tAv = 0$. I neglected to say above that the focal locus corresponds to the closest point on either side of $S$ along the geodesic normal to $x$ where this equation holds. Therefore, there is a focal point on the half line where $t > 0$ only if there is a negative principal curvature and the focal point is at distance $-1/\kappa$, where $\kappa$ is the negative principal curvature with largest magnitude. There is a focal point on the other half line only if there is at a positive principal curvature, and it is at distance $1/\kappa$, where $\kappa$ is the largest positive principal curvature.
This paragraph doesn't answer the question, but discusses some more elementary
related calculations (as in Hamilton's Nonsingular Solutions paper). Consider
a closed surface $\Sigma_{t}$ in a $3$-manifold $(M,g(t))$ evolving by Ricci
flow, where $\Sigma_{t}$ evolves in the normal direction $N$ with velocity
function $V$. Then the induced metric $\operatorname{I}_{t}$ on $\Sigma_{t}$
evolves by $\frac{\partial}{\partial t}\operatorname{I}_{t}=2\operatorname{II}
_{t}V-2\operatorname{Ric}|_{T\Sigma_{t}}$, where $\operatorname{II}_{t}$ is
the second fundamental form of $\Sigma_{t}$ and $\operatorname{Ric}
|_{T\Sigma_{t}}$ is the restriction of the Ricci tensor of $g\left( t\right)
$ to $T\Sigma_{t}$. Let $dA_{t}$ denote the area element of $\Sigma_{t}$. Then
$\frac{\partial}{\partial t}dA_{t}=\frac{1}{2}\operatorname{trace}
_{\operatorname{I}_{t}}(\frac{\partial}{\partial t}\operatorname{I}_{t}
)dA_{t}=(H_{t}V-R+\operatorname{Ric}(N,N))dA_{t}$, where $H_{t}$ is the mean
curvature of $\Sigma_{t}$. In particular, if $V=0$, then the area
$\operatorname{A}_{t}$ of $\Sigma_{t}$ evolves by $\frac{d}{dt}
\operatorname{A}_{t}=\int_{\Sigma_{t}}(-R+\operatorname{Ric}(N,N))dA_{t}
=\int_{\Sigma_{t}}(-\frac{1}{2}R-\operatorname{sect}(T\Sigma_{t}))dA_{t}$,
where $\operatorname{sect}(T\Sigma_{t})$ denotes the sectional curvature of
the plane $T\Sigma_{t}$. On the other hand, by the Gauss equations for
$\Sigma_{t}\subset M$, the intrinsic Gauss curvature of $(\Sigma
_{t},\operatorname{I}_{t})$ is $K_{t}=\operatorname{sect}(T\Sigma_{t}
)+\det(\operatorname{II}_{t})$. So $\frac{d}{dt}\operatorname{A}_{t}
=\int_{\Sigma_{t}}(-\frac{1}{2}R-K_{t}+\det(\operatorname{II}_{t}))dA_{t}$.
This formula is nice at some time, for example, if $\Sigma_{t}$ is a minimal
surface and the scalar curvature is bounded from below $R\geq-C_{t}$ (the
latter is indeed true for Ricci flow), when and where $\frac{d}{dt}
\operatorname{A}_{t}\leq\frac{C_{t}}{2}\operatorname{A}_{t}-2\pi\chi(\Sigma)$
since $\det(\operatorname{II}_{t})\leq0$ follows from $H_{t}=0$ and by the
Gauss-Bonnet formula.
The relevant computations must be somewhere in the literature, but I don't
know where; so the following is off the top of my head and needs to be
checked. About the normal $N_{t}$ in the case of a static hypersurface,
consider a family of inner products $g_{t}$ on a vector space $E$ (e.g.,
$T_{x}M$) and a fixed hyperplane $P$ (e.g., $T_{x}\Sigma$). By $g_{t}
(X,N_{t})\equiv0$ for each $X\in P$, we have $\frac{\partial g_{t}}{\partial
t}(X,N_{t})+g_{t}(X,\frac{\partial N_{t}}{\partial t})=0$. So $\frac{\partial
g_{t}}{\partial t}(N_{t})+g_{t}(\frac{\partial N_{t}}{\partial t})=cN_{t}$ for
some $c\in\mathbb{R}$, identifying $T_{x}M$ and $T_{x}^{\ast}M$ by $g_{t}$
here and below. Dotting with $N_{t}$ yields $c=\frac{1}{2}\frac{\partial
g_{t}}{\partial t}(N_{t},N_{t})$ since $g_{t}(\frac{\partial N_{t}}{\partial
t},N_{t})=-\frac{1}{2}\frac{\partial g_{t}}{\partial t}(N_{t},N_{t})$ from
$g_{t}(N_{t},N_{t})\equiv1$. We obtain $\frac{\partial N_{t}}{\partial
t}=\frac{1}{2}\frac{\partial g_{t}}{\partial t}(N_{t},N_{t})N_{t}
-\frac{\partial g_{t}}{\partial t}(N_{t})$. Combining this with some other
formulas of this nature, such as the standard $\frac{\partial}{\partial
t}\nabla_{t}$, where $\nabla_{t}$ denotes the Levi-Civita connection of
$g_{t}$, one should be able to compute the evolution of $\operatorname{II}
_{t}$, etc.
[Dec 3, 2013] In response to Chris Gerig's question:
Let $F:N\times(0,T)\rightarrow M$ be a parametrized hypersurface in a
Riemannian manifold $(M^{n},g)$. The first fundamental form (induced metric)
at time $t$ is $\operatorname{I}_{t}(X,Y)=g(dF_{t}(X),dF_{t}(Y))$, where
$F_{t}(x)=F(x,t)$. The unit normal $\nu$ and the velocity $dF_{t}
(\frac{\partial}{\partial t})=\frac{\partial F}{\partial t}=V\nu$ are vector
fields along the map $F$. We compute that
$$
\frac{\partial}{\partial t}\operatorname{I}_{t}(X,Y)=g(\frac{D}{dt}
dF_{t}(X),dF_{t}(Y))+g(dF_{t}(X),\frac{D}{dt}dF_{t}(Y)),
$$
where $\frac{D}{dt}$ is covariant differentiation along the path $\alpha
_{x}(t)=F(x,t)$. Basically, since $[\frac{\partial}{\partial t},X]=0$ in
$N^{n-1}\times(0,T)$ and by pushing this forward by $F$, we have $\frac{D}
{dt}dF_{t}(X)=\frac{D}{dX}\left( dF_{t}(\frac{\partial}{\partial t})\right)
=\frac{D}{dX}\left( V\nu\right) $, where $\frac{D}{dX}$ is covariant
differentiation along $F$ restricted to a path in $N\times\{t\}$ tangent to
$X$ (heuristically, $\nabla_{dF_{t}(\frac{\partial}{\partial t})}
dF_{t}(X)-\nabla_{dF_{t}(X)}dF_{t}(\frac{\partial}{\partial t})=[dF_{t}
(\frac{\partial}{\partial t}),dF_{t}(X)]=dF_{t}([\frac{\partial}{\partial
t},X])=0$). Using $\langle X,\nu\rangle=0$ and the product rule for $\frac
{D}{dX}$, we obtain
$$
\frac{\partial}{\partial t}\operatorname{I}_{t}(X,Y)=V\left( g(\frac{D}
{dX}\nu,dF_{t}(Y))+g(dF_{t}(X),\frac{D}{dY}\nu)\right) =2V\operatorname{II}
{}_{t}(X,Y)
$$
by the definition of the second fundamental form in terms of the derivative of
the unit normal.
Best Answer
Perhaps this—and its references both past & future ("cited by 152" subsequent papers)—will help...?
David (Xianfeng) Gu's work was cited by Deane Yang in the comments.
Here is one among the (many) later papers, coauthored by David Gu: