I am not entirely sure what they had in mind, but one possible interpretation of this is that you cannot "see" the difference between a non-reduced scheme and its induced reduced scheme structure geometrically, because they have the same set of points.
A non-reduced scheme is the same topological space as a reduced scheme with extra functions, so you have to take these extra functions into account and that is inherently algebraic.
In other words, you cannot expect to have a condition that says something about the dimension of the fibers or components (associated points) dominating the base like in criteria over reduced bases.
OK, let me try to give a more complex answer. Take the simple flatness criterion that
Let $f:X\to B$ be a morphism such that $B$ is integral, regular, and of dimension $1$. Then $f$ is flat if and only if every associated point of $X$ dominates $B$.
This is not a topological characterization, because it allows $X$ to have embedded points, yet it requires $B$ to be reduced. I suppose you could argue that the part of the condition of the theorem that applies to $B$ is indeed topological, but I would not expect anybody call this criterion topological.
Most other flatness criteria require something about the fibers of the map, which is again only topological on the base, but more than that on the total space and/or fibers.
Also, one might argue that when someone says topological with respect to an algebraic scheme they might mean the euclidean or the étale topology. In some sense to work with the Zariski topology is already a little bit geometric rather than topological.
I realize that this may not sound satisfactory to you, but I think many remarks authors make do not hide some big secret. If they had some complex picture in mind when they wrote it, I bet they would have said more. Then again, I do not claim to know what was/is on their mind. As Donu suggests you could ask the authors. If you do, please share the answer with us!!