Petri's theorem (see ACGH III.3) states that if $C\subset P^{g-1}$ is a canonical curve (i.e. the canonical image of a non hyperelliptic curve) of genus $g\ge 4$ then the ideal of $C$ is generated by quadrics iff $C$ is not trigonal or a plane quintic (for trigonal curves or plane quintics the intersection of all quadrics through $C$ is a surface).
If $C$ is hyperelliptic, the dimension of the space of quadrics through the canonical image $\Gamma$ is $(g+1)g/2-2g+1=(g-1)(g-2)/2$. The ideal of $\Gamma$ is also generated by quadrics.
If $g-2$ is not a power of 2, a trivial lower bound is $d\ge g-1$. Indeed by Bezout's theorem (see Fulton, Intersection theory, 8.4) if the intersection of $g-2$ quadrics of $P^{g-1}$ is a curve of degree $d$, then $d$ divides $2^{g-2}$.
The following $2g-5$ quadrics cut the rational normal curve in $P^{g-1}$ set theoretically: $x_0x_2-x_1^2,x_0x_3-x_1x_2, \dots, x_0x_{g-1}-x_{g-2}x_1, x_2^2-x_1x_3, x_3^2-x_2x_4,\dots,x_{g-2}^2-x_{g-3}x_{g-1}$.
I don't know if one can do better.
It is true that $Z$ spans $L$ — even if $X$ isn't ACM. You can also allow $X$ to be singular (but you do need $X$ irreducible and non-degenerate, of course). To illustrate one of the main ideas it is useful to first look at the case when $X$
is a curve.
If $X$ is a curve. Let $M$ be the span of $Z$ and suppose that $M\neq L$. (In the curve case, $L$ will be a hyperplane). Let $p$ be any point of $X$ outside of $Z$ and let $H$ be any hyperplane containing $M$ and $p$. Then $H\cap X$ contains at least $d+1$ points, so by Bezout's theorem the intersection cannot be zero dimensional. Since $X$ is irreducible and one dimensional, this means that the intersection must be all of $X$, so $X$ is contained in $H$, contrary to hypothesis.
The general case.
The idea when $k\geqslant 2$ is to show that if $H$ is a general hyperplane containing $L$ then $H \cap X$ is irreducible and non-degenerate (i.e, the intersection $H\cap X$ is not contained in a smaller linear space of $H$). But now all dimensions have been reduced by $1$, and so iterating this procedure reduces us to the curve case, which we've already solved.
To set this up, note that hyperplanes in $\mathbb{P}^n$ containing $L$ are parameterized by a $\mathbb{P}^{k-1}$ (If $V$ is the underlying vector space of $\mathbb{P}^{n}$, $W$ the underlying vector space of $L$, then the hyperplanes are parameterized by the projectivization of $(V/W)^{*}$). We'll use $H$ to refer both to a point of $\mathbb{P}^{k-1}$ and the corresponding hyperplane in $\mathbb{P}^n$ containing $L$. Define $\Gamma\subset \mathbb{P}^{k-1}\times (X\setminus Z)$ to be the set
$$\Gamma = \left\{(H,p) \mid p\in H\right\}$$
i.e, the pairs $(H,p)$ so that $H$ is a hyperplane containing $L$, and $p$ a point of $H\cap X$ not on $Z$.
If we fix $p$, then the set of possible $H$'s satisfying this condition are simply the hyperplanes $H$ containing the span of $L$ and $p$, and this is parameterized by a $\mathbb{P}^{k-2}$. In other words, $\Gamma$ is a $\mathbb{P}^{k-2}$ bundle over $X\setminus Z$. (This fibration is where we use $k\geqslant 2$.) Since $X\setminus Z$ is irreducible this implies that $\Gamma$ is irreducible.
Let $\overline{\Gamma}$ be the Zariski-closure of $\Gamma$ in $\mathbb{P}^{k-1}\times X$. Then $\overline{\Gamma}$ is irreducible since $\Gamma$ is. For a fixed $H\in \mathbb{P}^{k-1}$ the fibre of the projection $\overline{\Gamma}\longrightarrow \mathbb{P}^{k-1}$ over $H$ is simply the intersection $X\cap H$, of dimension $k-1$.
Now let $q$ be any point of $Z$. Then $q\in X\cap H$ for every $H\in \mathbb{P}^{k-1}$ so $q$ gives a section of $\overline{\Gamma}\longrightarrow\mathbb{P}^{k-1}$. Since $Z$ consists of $d$ distinct points where $d$ is the degree of $X$ we conclude that $q$ is a smooth point of $X$. Finally, since $Z$ is the intersection of all $X\cap H$ for $H\in \mathbb{P}^{k-1}$ this implies that the general intersection $X\cap H$ is smooth at $q$. Summarizing, we have a section of the map which generically lies in the smooth locus of the fibres. Since $\overline{\Gamma}$ is irreducible, this implies that the generic fibre is irreducible, i.e, if $H$ is a generic hyperplane containing $L$, then $H\cap X$ is irreducible.
(The intuitive reason for this implication is that, generically over $\mathbb{P}^{k-1}$ the section lets us pick out precisely one irreducible component of the fibre. The union of these components gives us a subset of $\overline{\Gamma}$ which has the same dimension as $\overline{\Gamma}$, and hence whose closure must be all of $\overline{\Gamma}$ by irreducibility. But if there is more than one component in a general fibre, this is a contradiction, thus the general fibre must be irreducible. To make this intuitive construction rigorous requires passing to the normalization of $\overline{\Gamma}$ and then looking at the Stein factorization of the map from the normalization to $\mathbb{P}^{k-1}$. The section gives a generic section of the finite part of the Stein factorization, and that allows one to construct the ``union of the components containing the section''.)
Finally, the same trick as in the curve case also shows us that for any hyperplane $H$, $H\cap X$ must be non-degenerate. Let $Y=H\cap X$, so that $Y$ is a variety of degree $d$ and dimension $k-1$. Let $M$ be the span of $Y$. If $M\neq H$ then pick any point $p\in X\setminus Y$ and let $H'$ be any hyperplane containing $M$ and $p$. Then $H'\cap X$ can't be all of $X$ (since this would contradict the non-degeneracy of $X$), so $Y'=H'\cap X$ must be a subvariety of dimension $k-1$ (more precisely, all components of $Y'$ have dimension $k-1$) and degree $d$. But $Y$ is therefore a component of $Y'$, and the equality of degrees tells us that $Y'$ can't have any other components so we must have $Y'=Y$. This contradicts the fact that $p\in Y'$ and $p\not\in Y$.
Together this shows the required inductive step: If $H$ is a general hyperplane containing $L$ then $H\cap X$ is irreducible and non-degenerate.
Other remarks. I'm guessing from the setup of the question that you want to apply the result for a particular $L$ that you have chosen. If, in the application, you're allowed to pick a general $L$ then you can say something stronger. The classical uniform position principle (where ''classical'' in this case means ''established by Joe Harris in the 80's'') states that for a general subspace $L$ of dimension $n-k$ the finite set of $d$-points in $Z=L\cap X$ have the property that any subset of $r+1$ of the points (with $r\leqslant n-k$) span a $\mathbb{P}^{r}$. Picking $r=n-k$, this means that any subset of $n-k+1$ points of $Z$ spans all of $L$, and so in particular $Z$ spans $L$. (Note that $d\geqslant n-k+1$; for instance, as a consequence of the argument above: if $d < n-k+1$ then the $d$ points of $Z$ would never span $L$.)
Best Answer
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.