The Sylvester Gallai Theorem and Sections of Varieties with "Simple Topology".
There are many cases of the question as stated that follow quickly from the standard Sylvester-Gallai theorem. If $V$ is an $r$-dimensional variety, then its intersection with a generic $(n-r)$-plane is a finite set of points. You can then apply the standard Sylvester-Gallai theorem, or the high-dimensional generalization stated here.
There are cases where nothing can be said for singular varieties. As a warm-up, let's consider a set which is not an algebraic variety but a union of line segments. Then it could be the union of all of the interior diagonals of a convex polytope $P$ with complicated facets. For instance you could take all of the interior diagonals of the Cartesian product of two $n$-gons. Any 3-plane that intersects $P$ 3-dimensionally has to intersect many of the edges.
A line segment is not a real algebraic variety. However, it can be replaced by a thin needle with cusps at the ends that is a real algebraic variety. You can replace all of the diagonals with these needles, as long as you skip the edges of $P$ itself, and the result will still lie in the convex hull of $P$.
A needle of this type can have a cross-section of any dimension and very complicated topology. If you asked for a hyperplane that specifically intersects in more than a finite set, then the diagonal-needle construction can force a lot of topology.
You could specifically look at non-singular varieties. I don't have a rigorous result here, but the smooth restriction would make it difficult to avoid hyperplanes that do something at the boundary of the convex hull of $V$. The Sylvester-Gallai theorem is more about things that have to happen in the interior if they do not happen at the boundary of the convex hull.
You could bound the degree of the variety $V$. Then a simple compactness argument bounds the complexity of intersects, and there are a lot of interesting bounds on the topology of $V$ itself. But that also goes against the spirit of Sylvester-Gallai, because the number of points in that result is not bounded.
Maybe a more interesting variation is to keep a finite intersection, but replace the hyperplane with a $V$ with bounded degree. However, that is no longer the question posted.
The question is a bit open-ended. I can think of several constructions that seem to ask for a less open-ended question, or a question which is open-ended in a different way.