So, I have the following definition of $\mathbb{P}_A^n$ for an arbitrary (commutative) ring $A$, from Hartshorne:
Set $S=A[x_0,\ldots,x_n]$, so that $S=\bigoplus_{d\geq 0}S_d$ as a graded ring, $S_+=\bigoplus_{d\geq 1}S_d$, and for convenience, let $S^\mathrm{H}=\bigcup_{d\geq 0} S_d$ denote the homogeneous elements. We define the set $\mathrm{Proj}\ S=\{\mathfrak{p}\subset S \mid \mathfrak{p} \mathrm{\ hmg.\ prime}, S_+ \nsubseteq\mathfrak{p}\}$ with closed sets $V(\mathfrak{a})=\{\mathfrak{p}\in\mathrm{Proj}\ S\mid \mathfrak{p}\supseteq\mathfrak{a}\}$ for all homogeneous ideals $\mathfrak{a}\subseteq S$.
Next, for all $\mathfrak{p}\in\mathrm{Proj}\ S$, we set $T_\mathfrak{p}=S^\mathrm{H}\setminus\mathfrak{p}$, $S_{(\mathfrak{p})}=\{\frac{f}{g}\in T_\mathfrak{p}^{-1} S \mid f\in S^\mathrm{H}, \deg f = \deg g\}$. Finally, for any open subset $U\subseteq\mathrm{Proj}\ S$, we define $$\mathcal{O}(U)=\{s:U\to\bigsqcup_\mathfrak{p} S_{(\mathfrak{p})} \mid \forall\ \mathfrak{p}\in U, s(\mathfrak{p})\in S_{(\mathfrak{p})}, \exists\ \mathfrak{p}\in V\subseteq U \mathrm{\ open}, a,f \in S^\mathrm{H} \mathrm{\ s.t.\ } \deg\frac{a}{f} = 0, \mathrm{\ and\ }\forall\ \mathfrak{q}\in V, f\notin\mathfrak{q}, s(\mathfrak{q})=\frac{a}{f}\in S_{(\mathfrak{q})}\}$$
When $A$ is an algebraically closed field, I can see the analogy with the projective space $\mathbb{P}^n$ of classical algebraic geometry. But for arbitrary rings, even for the simplest case of $A=\mathbb{Z}$, I struggle to make sense of this mess of symbols. Is there some good intuition to keep in mind when working with $\mathbb{P}_A^n$, or some simpler way of describing the ring of regular functions?
At the very least, I'd like to understand what's going on in $\mathbb{P}_\mathbb{Z}^n$, to gain some intuition for the more general case.
I agree that the point-set description is not very enlightening (the ratio of complexity of explicit description against complexity of abstract meaning can get much worse, e.g. Does the category of locally ringed spaces have products?); although one has to prove existence somehow, I think the right way to view the projective space is to look at the functor that the projective $n$-space represents, i.e. to describe how morphisms $X\to {\mathbb P}^n_A$ can be constructed:
This is a very intuitive thing: Roughly, given a point $x$ of $X$, you pick some $s_i$ not vanishing at $x$, and map $x$ to $(\frac{s_j(x)}{s_i(x)})$. Choosing a different $s_{i^{\prime}}$ modifies the tuple by the constant $\frac{s_i(x)}{s_{i^{\prime}}(x)}$, hence defines the same point in projective space. I like this idea of 'measuring' the $s_j(x)$ against some chosen nonzero $s_i(x)$, and to remove the ambiguity by factoring out constants. Conversely, any morphism $p: X\to {\mathbb P}^n_A$ determines ${\mathcal L} := p^{-1}({\mathscr O}(1))$, and the $s_i$ are the pullbacks of the standard sections of ${\mathscr O}(1)$.
This description simplifies to the classical description of projective spaces when restricted to $X$ which does not admit any nontrivial line bundles: Then ${\mathcal L}\cong {\mathscr O}_X$ and a map $X\to {\mathbb P}^n_A$ is a tuple $(f_0,...,f_n)\in {\mathscr O}_X^n$ of generators of ${\mathscr O}_X^{n+1}$, modulo the action of ${\mathscr O}_X(X)^{\times}$. Taking $X=A=\text{Spec}(k)$ for example, this gives ${\mathbb P}^n_k(k)=(k^{n+1}\setminus\{0\})/k^{\times}$.
In the midway, you might consider $X=\text{Spec}(B)$ for some $A$-algebra $B$. Then specifying a map $X\to {\mathbb P}^n_A$ comes down to specifying a projective $B$-module $P$ together with a surjection $B^{n+1}\twoheadrightarrow P$, up to isomorphism. For $B=A={\mathbb Z}$ (over which every projective module is free) you get ${\mathbb P}^n_{\mathbb Z}({\mathbb Z})=\text{coprime}({\mathbb Z}^{n+1})/\{\pm 1\}$, where $\text{coprime}({\mathbb Z}^{n+1})$ is the set of $(a_0,...,a_n)$ which have no common prime divisor.
Note: You can also see the paving of ${\mathbb P}^n_A$ by ${\mathbb A}^n_A$ this way: For any $i\in\{0,1,...,n\}$ you have the subfunctor restricting to those $({\mathcal L},s_0,...,s_n)$ such that $s_i$ is nowhere vanishing. But if you have a nowhere vanishing section of a line bundle, this section trivializes the bundle, so the datum essentially comes down to specifying $n$ global functions on $X$, which is the same as a morphism $X\to {\mathbb A}^n_A$.