On page 15 of this manuscript, the author says the following on why lines can be represented by 2-blades:
I'm facing trouble understanding the point the author is trying to make in the second paragraph. They call the equation just presented an "inheritance property", and I'm not sure why. Also, they assert that that equation, no matter how points are in fact represented, is enough justification for lines to be represented as 2-blades, but I don't see why that would be the case. It's true that the equation holds for vectors, but it fails for higher-grade blades.
An immediate case is that of two orthogonal bivectors $A$ and $B$ in three-dimensional space. Their inner product $A \cdot B$ is zero, but the inner product of $A$ by vector factors that make up $B$ are not all zero, since both bivectors share a common vector. So it would seem that if points were to be represented by anything other than vectors, this equation alone shouldn't really justify having lines being represented as 2-blades.
$ \newcommand\lcontr{\mathbin\rfloor} \newcommand\R{\mathbb R} $You are correct to be confused as there is a lot wrong with this passage. I will focus on the Euclidean case, but we can do much the same with any metric signature.
I think it would be most expedient to just describe the whole setup correctly:
In "usual" geometric algebra we represent a flat through some chosen origin as a blade $X$, and we do this by interpreting vectors $p$ as lines; $X$ represents the subspace $[X]$ spanned by all $[p]$ such that $p\wedge X = 0$. It then follows that $X\wedge Y = 0$ if the corresponding subspaces contain a common line; otherwise $[X\wedge Y]$ is the subspace spanned by $[X]$ and $[Y]$. Let's call this setup LGA (Linear Geometric Algebra).
This is consistent with the interpretation of vectors as points since any point of $[p]$ is a multiple of $p$, however this way of representing points will not carry through in the following.
There is a dual interpretation made by the following substitutions to the above: $$ \text{line} \mapsto \text{hyperplane},\quad \text{contains} \mapsto \text{is contained in},\quad \text{span} \mapsto \text{intersection}. $$ This gives us the following text: we represent a flat through some chosen origin as a blade $X$, and we do this by interpreting vectors $p$ as hyperplanes; $X$ represents the subspace $[X]$ which is the intersection of all $[p]$ such that $p\wedge X = 0$. It then follows that $X\wedge Y = 0$ if the corresponding subspaces are contained in a common hyperplane; otherwise $[X\wedge Y]$ is the intersection of $[X]$ and $[Y]$. Let's call this setup LGA*.
This is all independent of metric. If we do have a nondegenerate metric, then we can interpret an LGA as an LGA* via pseudoscalar multiplication. The vector metric $v\mapsto v^2$ induces a pseudovector metric $V \mapsto \widetilde VV$, and this relationship is reciprocal: an $(n,0,0)$ vector metric induces an $(n,0,0)$ pseudovector metric. In this sense the difference between LGA and LGA* is immaterial. Notice that we can import the LGA* exterior product into LGA (or vice-versa), yielding the regressive product defined by some variation of $$ X\vee Y = (XI)\wedge(YI)\,I^{-1} = (XI)\lcontr Y. $$ (This left contraction $\lcontr$ is much like your dot product but better behaved on edge cases; the above equations are valid for any two multivectors.) In particular in LGA* we can find out which pseudovector lines $l$ (and hence points) make up a blade $X$ via $l\vee X = 0$; if we take a vector $v$ such that $l = vI^{-1}$ then we have $v\lcontr X = 0$. This is perhaps what the passage you cite has in mind. Notice that we are mixing LGA and LGA* interpretations here. You may also have heard of "outer product nullspace" and "inner product nullspace" interpretations; this is where the inner product nullspace idea really comes from.
This would be all well and good, except it completely fails in PGA. Later I will show that PGA can be constructed canonically as a GA* with metric $(n,0,1)$. But for now take my word that PGA is projectived at the cost of an extra dimension so that we can represent arbitrary flats rather than just those through a fixed point, and that vectors must represent hyperplanes. Once we make this determination, we then have to confront the fact that the symmetry highlighted in the previous paragraph has been broken, making a PGA* with vectors as points a fundamentally different entity from PGA. I urge you to calculate for yourself that an $(n,0,1)$ signature on vectors induces a $(1,0,n)$ signature on pseudovectors, and then that $(1,0,n)$ induces $(0,0,n+1)$. You can also determine that pseudoscalars are never invertible, making the previous definition of regressive product and the "inner product nullspace" untenable. In fact, look at what happens when we do try to represent points with vectors and "inner product nullspaces": noting that $(v+w)\lcontr X = v$ whenever $w^2 = 0$, signature $(1,0,n)$ is utterly broken since there is only one degree of freedom, and signature $(n,0,1)$ gives us a nonunique/basis dependent representation of points. The latter is probably the intent of the passage you cite, but maybe like me this approach leaves a bad taste in your mouth. (More precisely it depends on a choice of origin; not a great thing to have for a construction whose purpose is to free us from an origin!)
Despite this break in symmetry, we can still determine the pseudovector point content of a blade $X$ because the point content is a non-metric property. We simply need a way to translate between PGA and PGA*; this can be done (almost) canonically using the Poincaré isomorphisms, which I will not describe here (but are intimately tied to the pseudoscalar multiplication we used previously). With this in hand, we can define the regressive product and recover the point content as all pseudovectors $P$ such that $P\vee X = 0$.
Construction of PGA
(And actually CGA, too!) Let $V$ be an $n$-dimensional real vector space with Euclidean metric. A cycle is a function $P : V \to \R$ of the form $$ P(x) = ax^2 + b\cdot x + c $$ for scalars $a,c$ and vector $b$. The possible solution sets of $P(x) = 0$ are precisely any point, $(n-1)$-sphere, or hyperplane (or the empty set). The set of all cycles $V^*_2$ forms a vector space. We may put $P(x) = 0$ in the following form by completing the square: $$ (2ax + b)^2 = b^2 - 4ac. $$ The RHS is a quadratic form $Q$ on $V^*_2$ with signature $(n,2,0)$. $P$ is a sphere or hyperplane when $Q(P) \ne 0$ and is a point when $Q(P) = 0$ (or an ideal element like the hyperplane-at-infinity $P(x) = 1$.) The geometric algebra of $Q$ over $V^*_2$ is precisely CGA.
Now keep the same metric but restrict $V^*_2$ to the subspace of 1-cycles $V^*_1$ which have the form $P(x) = b\cdot x + c$. These are precisely the cycles which represent hyperplanes. They can be intrinsically defined as a cycle $P$ which is a constant function (a 0-cycle) or for which there is some constant $c$ such that $x \mapsto P(x) - c$ is linear. The restriction of $Q$ to $V^*_1$ has signature $(n,0,1)$; the geometric algebra of $Q$ over $V^*_1$ is PGA.