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.
Under one additional condition, the answer to this problem is affirmative.
The proof involves the following implication of the Affine Desargues Axiom:
The Affine Moufang Axiom: for every parallel lines $A,B,C$ in an affine plane $(X,\mathcal L)$ and any points $a,a'\in A\setminus (B\cup C)$, $b,b'\in B\setminus(A\cup C)$, $c,c'\in C\setminus(A\cup B)$, if $\overline{ab}\parallel\overline{a'b'}$ and $\overline{bc}\parallel \overline{b'c'}$, then $\overline{ac}\parallel \overline{a'c'}$.
The implication of the Affine Moufang Axiom (called the little Desargues Theorem) from the Affine Desargues Axiom (called the Affine Desargues Theorem) is discussed in this MO-post.
Also we shall use the transitivity of the parallelity relation in the affine planes.
Proposition. Let $A,B,C$ be three lines in an affine plane $(X,
\mathcal L)$. If $A\parallel B$ and $B\parallel C$, then $A\parallel C$.
Proof. If $A\nparallel C$, then the lines $A,C$ have a common point $b$. If $b\in B$, then $A\parallel B\parallel C$ implies $A=B=C$ and hence $A\parallel C$. So, $b\notin B$ and $A=C$ by the axiom 4 of an affine plane. $\quad\square$
Now we can prove the partial answer to the MO-problem.
Theorem. Let $a,b,c,d$ be distinct points of a Desarguesian affine space $(X,\mathcal L)$ and $p,q,r,s\in X$ be points such that $p\in\overline{ab}$, $q\in\overline{bc}$, $r\in\overline{cd}$ and $s\in\overline{ad}$. If the lines $\overline{ca}$, $\overline{cb}$, $\overline{cd}$ are distinct and $\overline{pq}\parallel \overline{ac}$, $\overline{qr}\parallel \overline{bd}$, $\overline{rs}\parallel \overline{ac}$, then $\overline{ps}\parallel \overline{bd}$.
Proof. It follows from $\overline{qr}\parallel \overline{bd}$ and $q\ne r$ that $q\ne c\ne r$ and hence $q\notin\overline{ac}$. By the axiom 4 of an affine space, there exists a unique point $t\in \overline{ac}$ such that $\overline{ab}\parallel \overline{qt}$. Since $\overline{tq}\parallel\overline{ab}$ and $\overline{qr}\parallel\overline{bd}$, the Affine Desargues Axiom guarantees that $\overline{tr}\parallel \overline{ad}$. Since $\overline{pq}$, $\overline{at}$, $\overline{sq}$ are three parallel lines, the Affine Moufang Axiom guarantees that $\overline{qr}\parallel \overline{ps}$ and hence $\overline{ps}\parallel \overline{bd}$, by the transitivity of the parallelity relation.
$\square$
Remark: If $\overline{ca}=\overline{cb}\ne\overline{cd}$, then it is easy to find points $p,q\in \overline{ca}=\overline{cb}$, $r\in\overline{cd}$, and $s\in\overline{ad}$ such that $\overline{qr}\parallel \overline{bd}$ and $\overline{rs}\parallel\overline{ca}$, but $\overline{ps}\nparallel \overline{bd}$.
Best Answer
For the solution see the attached figure here. It is the counterexample with $n=8$. Vertical lines represent all possible vertical lines which have less that $n$ intersection points with given lines.
How I found this counterexample?
By duality of the real projective plane, the problem is equivalent to the following:
Given $n$ distinct points on the plane such that they are not collinear and no two of them lie on the vertical line, is it true that there exists a direction $\mathcal{D}$ such that when passing a line from $\mathcal{D}$ through every given point we obtain $n-1$ or $n-2$ distinct lines?
And the answer is no. The counterexample consists of the vertices of the regular octagon rotated so that no two vertices lie on the vertical line. The figure corresponds to such octagon.
More notes. After the discussion with my colleagues from Katowice we concluded a little more about the problem.
Concerning the equivalent (dual) problem, given a set of $n$ points on the plane, we may consider the set $I$ of all natural numbers $j$ such that there exists a direction $\mathcal{D}$ yielding exactly $j$ distinct lines (when passing a line from $\mathcal{D}$ through every given point).
For a regular polygon with $n=2k$ vertices we have $I=\{k,k+1,2k\}$ and for a regular polygon with $n=2k+1$ vertices we have $I=\{k+1,2k+1\}$. Therefore $n=7$ is sufficient for a counterexample to the original problem.
One more observation is that transforming a regular polygon through an affine bijection of $\mathbb{R}^2$ preserves parallel lines, so the image of $n$ points under such transformation has the same set $I$ as the original points.