This comes from Friedman's PDE of parabolic type page 69
Let $D$ be a domain.and $S$ be its boundary "on the side". i.e. the boundary not including the intitial condition.
We shall say that $S$ has outside strong sphere property if for every $Q=(x^0,t^0)\in S$ there exists a ball $K$ with centre $(\bar{x},\bar{t})$ such that $K\cap \bar{D} = \{Q\}$ and
$|\bar{x}-x|\geq \mu(Q)>0$ for all $(x,t)\in\bar{D}$, $\ \ \ |t-t^0|<\epsilon$,
where $\epsilon$ is independent of $Q$
I do not quite understand what this mean. It says for every $Q$ on $S$ there is a ball such that the centre of the ball is $(\bar{x},\bar{t})$
This is an exterior sphere condition, with a constraint on the "direction" from which the sphere touches.
The condition $K\cap \bar{D} = \{Q\}$ means that the sphere $K$ sits completely outside of $\bar{D}$, except for the point $Q$ which lies on the boundary of $D$ and the boundary of the ball $K$. Thus $K$ touches $D$ from the exterior (draw a picture).
The other condition says that the vector pointing from the center of $K$ to the point $Q$ is not too much aligned with the $t$ direction, so the ball touches from the "side" of $D$, and not from the "top" or "bottom". For instance, if $x^0=\bar{x}$, then the condition cannot hold because you can take $(x,t)=(x^0,t^0)=Q$. This is the case where $v$ is entirely in the $t$ direction, so the ball touches from above or below.