I am reading 'A Royal Road to Algebraic Geometry' by Audun Holme and in 1.1 he mentions that by diving $\mathbb{P}^k_n$, the projective $n$-space space over a field $k$, over and over, we can eventually arrive at
$$ \mathbb{P}_k^n=\mathbb{A}_k^n\cup\mathbb{A}_k^{n-1}\cup\cdots\cup\mathbb{A}_k^1\cup\mathbb{P}_k^0, $$
where $\mathbb{A}_k^n$ is the affine $n$-space over the field $k$.
I understand the derivation but I have no idea how to think about this set. I also don't currently see the link between the decomposition for the set above and thinking about $\mathbb{P}_k^n$ as the lines in $\mathbb{A}_k^{n+1}$ that pass through the origin.
Any help is much appreciated - I am a beginner to this field.
Kind regards.
For an intuitive understanding, it's best to think about the case where the field $k$ is $\mathbb{R}$, and $n=2$. That way we can visualize everything.
First, the elements of $\mathbb{P}_\mathbb{R}^2$ are lines through the origin in $\mathbb{R}^3$. Each of these intersects the sphere $\mathbb{S}^2$ in two points. We can regard $\mathbb{P}_\mathbb{R}^2$ as these pairs of points. That is, each projective point is an identified pair of antipodal points on the sphere. (I won't bother with the $\mathbb{R}$ subscript from now on.)
So the decomposition you mention is closely related to the decomposition $$\begin{align*}\mathbb{S}^2 &=2\mathbb{D}^2\sqcup\mathbb{S}^1\\ &=2\mathbb{D}^2\sqcup 2\mathbb{D}^1\sqcup\mathbb{S}^0 \end{align*}$$ where on the first line, $2\mathbb{D}^2$ indicates the two hemispheres and $\mathbb{S}^1$ the equator. Then on the next line, the $\mathbb{S}^1$ equator is decomposed into two (open) semicircles, and $\mathbb{S}^0$ which is two points.
Once we start identifying antipodal pairs, $\mathbb{S}^2$ becomes $\mathbb{P}^2$ and $2\mathbb{D}^2$ becomes just $\mathbb{D}^2$: we can discard the lower hemisphere (say). Likewise for $2\mathbb{D}^1$.
Now, this is a topological decomposition. For a geometric decomposition, we note that the correspondence between part of $\mathbb{P}^2$ and the upper hemisphere (say) was obtained by intersecting the lines through the origin with the sphere. But we could also intersect with a plane tangent to the hemisphere at the north pole. That gives you $\mathbb{A}^2$. Likewise, for one of the semicircles, we can do the intersections with a line tangent to it at its midpoint. So every projective point of $\mathbb{P}^2$ is replaced with a "representative" point on the plane, the line, or the sole point in $\mathbb{P}^0$.
Hope this helps!