I am trying to understand the part of this answer explaining why $A_8\cong\mathrm{PSL}_4(\mathbb{F}_2)$.
Let $|X|=8$. We can form the free vector space $\mathbb{F}_2X=\mathbb{F}_2^8$ with the usual dot product and consider the subquotient $v^{\perp}/v$ where $v=(1,\cdots,1)$. Or equivalently, the vector space $(\mathcal{P}(X),\Delta)$ where $\mathcal{P}(X)$ is the power set and $\Delta$ is symmetric difference with bilinear form $|A\cap B|\bmod 2$, then consider $X^{\perp}/X$. This is a $6$-dim vector space with subset $Q=\binom{X}{4}$ of unordered quadruples (mod complement) with $\frac{1}{2}\binom{8}{4}=35$ elements. The action of $S_8$ on $X$ induces one on $Q$.
On the other hand, the set of $2$-subspaces of $\mathbb{F}_2^4$, i.e. the Grassmanian $\mathrm{Gr}_2(\mathbb{F}_2^4)$, also has $35$ elements, it presumably has some kind of embedding in $\Lambda^2\mathbb{F}_2^4$ (which mods by $x\wedge x=0$, so is $6$-dim spanned by $e_i\wedge e_j$), the projectivized form is called the Klein quadric (which is presumably the same because $\mathbb{F}_2$), and carries an action of $\mathrm{PGL}_2(\mathbb{F}_2)$ (which is the same thing as $\mathrm{PSL}_2(\mathbb{F}_2$) and $\mathrm{GL}_4(\mathbb{F}_2)$ and $\mathrm{SL}_4(\mathbb{F}_4)$ because again, $\mathbb{F}_2$).
Supposedly there is a bijection $Q\leftrightarrow\mathrm{Gr}_2(\mathbb{F}_2^4)$ which makes the action of $A_8$ and $\mathrm{PSL}_4(\mathbb{F}_2)$ match, but I'm not seeing what it's supposed to be. What is this bijection, or in other words how is $\mathbb{F}_2^4$ related to $X$? Presumably there is some kind of functorial construction of $\mathbb{F}_2^4$ out of $X$ or vice-versa. Or are some arbitrary choices necessary?
From there, maybe I can see how $X=\mathbb{P}(\mathbb{F}_7)$ makes $\mathrm{PSL}_2(\mathbb{F}_7)$ relevant, and what the planes and lines business is in the so-called Klein correspondence.
Also - why $\mathrm{PSL}_2(\mathbb{F}_7)$ not $\mathrm{PGL}_2(\mathbb{F}_7)$? Why $A_8$ not $S_8$? Also curious if $\mathrm{Aff}_3(\mathbb{F}_2)=\mathbb{F}_2^3\rtimes\mathrm{GL}_3(\mathbb{F}_2)$ fits in somewhere, since it is a subgroup of both $S_8$ (with $X=\mathbb{F}_2^3$) and $\mathrm{PGL}_4(\mathbb{F}_2)$.
Edit. Some further thoughts.
Given a six-element set $Y$, we can take $Y^{\perp}/Y$ from $(\mathcal{P}(Y),\Delta)$ to get $S_6\to\mathrm{Sp}_4(\mathbb{F}_2)$ (with some set-theoretic skew form I can't remember), which is an isomorphism. We can embed $S_6$ in $A_8$ (multiply the odd elements of $S_6$ by $(78)$) and try to extend the action to $X^{\perp}/X$, but it doesn't seem like $Y^{\perp}/Y$ embeds naturally into it in the first place.
The sum $1+28+35$ seems to show up in places. First off, in the sizes of the $S_8$-orbits of $X^{\perp}/X$, in which case the sum is $\binom{8}{0}+\binom{8}{2}+\frac{1}{2}\binom{8}{2}$. Secondly, it appears in the $\mathrm{GL}_4(\mathbb{F}_2)$-orbits of $\Lambda^2\mathbb{F}_2^4$, comprised of elements of the form $a\wedge b+c\wedge d$ and $a\wedge b$. Thirdly for some reason it seems to show up in $\mathfrak{h}_3(\mathbb{F}_2)$, the $3\times 3$ Hermitian matrices over $\mathbb{F}_2$, as those nonzero elements with determinant $0$ or $1$ (so, presumably as the $\mathrm{GL}_3(\mathbb{F}_2)$-orbits). I can detail how I calculated the sizes of all these orbits if someone wants.
Is there some relationship between $\mathfrak{h}_3(\mathbb{F}_2)$ and $\Lambda^2\mathbb{F}_2^4$? Note $\Lambda^2\mathbb{F}_2^4$ naturally coincides with the diagonal-free $4\times4$ Hermitian matrices, say $\mathfrak{h}_4'(\mathbb{F}_2)$; is there a relationship between $\mathfrak{h}_3(\mathbb{F}_2)$ and $\mathfrak{h}_4'(\mathbb{F}_2)$? I imagine if there is, it has something to do with the exceptional homomorphism $\mathrm{SO}_4\to\mathrm{SO}_3$. (Note Hermitian and skew-Hermitian are equivalent over $\mathbb{F}_2$).
There is a Hodge-star operator $\ast$ on $\Lambda^2\mathbb{F}_2^4$; on $\mathrm{Gr}_2(\mathbb{F}_2^4)$ it replaces a $2$-subspace with its orthogonal complement. There are three $2$-subspaces which are fixed points, $\mathrm{span}\{e_i+e_j,e_k+e_\ell\}$ for the three partitions $\{\{i,j\},\{k,\ell\}\}$ of $\{1,2,3,4\}$. The 3D invariant subspace $(\Lambda^2)^{\ast}$ is spanned by the three linearly independent elements $e_i\wedge e_j+e_k\wedge e_\ell$, and $\ast$ acts trivially on $\Lambda^2/(\Lambda^2)^{\ast}$. Is there some corresponding involution $\ast$ on $X$ or $\mathfrak{h}_3(\mathbb{F}_2)$?
I'm aware there's some way to set up a vector space $\mathbb{F}_2^4$ using "even" and "odd" subgroups of $A_8$, but while this seems like a useful general trick to know I was hoping for something a bit less "pull-up-by-bootstraps" than looking at subgroups.