Dear Clark,
This answer is addressed primarily at the final paranthetical comment. There is a "dual" geometric description of $(\mathfrak g, K)$-modules to the Beilinson--Bernstein picture,
using orbits of the real group $G(\mathbb R)$ on the flag variety. It is employed by
Schmid and Vilonen in many of their papers, and there is an expository article about it
by Vilonen in one of the Park City proceedings.
In this dual picture, when one takes sections of the sheaves, one really gets $G(\mathbb R)$-representations; I'm not sure off the top of my head which models you get though (the smooth ones or some other ones, or whether you get an option depending on the particular
brand of sheaves you use).
Regarding your broader question, I've always imagined that one can find a model for a $(\mathfrak g, K)$-module using any brand of regularity you choose (analytic, smooth, distributional, hyperfunctions, and other brands in between, whatever they might be (and perhaps answering that is part of the point of the question!)), and that these will be ordered in the obvious way.
But this is certainly not the precise answer you want, and is only a vague intuition.
YCor gave a brief explanation in the comments. Based on your comment about seeking intuition, I think it would be helpful to explain concretely the backstory to the calculation YCor did:
Given any kind of arithmetically-defined group $\Gamma$, you want to find a real algebraic group / Lie group $G$ in which $\Gamma$ is arithmetic. For this, a necessary condition is that $\Gamma$ is discrete in $G$, but it's not sufficient. If $\Gamma \subset H \subset G$, then $\Gamma$ is not arithmetic in $G$ unless $G/H$ is compact, so we want to take $G$ as small as possible containing $\Gamma$.
Making $\Gamma$ discrete is easy. You just have to express elements of $\Gamma$ by integer coordinates, since $\mathbb Z^n$ is discrete in $\mathbb R^n$, so any subset of $\mathbb Z^n$ is discrete in any subspace of $\mathbb R^n$ containing it.
For example, for $\Gamma = SL(2, \mathcal O)$ or $\Gamma$ any subgroup of $SL(2, \mathcal O)$ (in, for simplicity, the special case $\mathcal O = \mathbb Z[\sqrt{a}]$), we can write any matrix $g$ as $M + N \sqrt{a}$ where $M, N $ are $2 \times 2$ matrices over the integers. This embeds $SL(2,\mathcal O)$ into $\mathbb Z^8$. For an arbitrary group over an arbitrary ring of integers $\mathcal O$ you can do the same thing, you just need to pick a $\mathbb Z$-basis of $\mathcal O$. So $SL_2 (\mathcal O)$ is discrete inside the manifold $\mathbb R^8$ parameterizing pairs $M ,N$ of $2\times 2$ matrices over the reals.
Of course we care about $\Gamma$ as a group and not just an abstract set. So you need to find a polynomial formula for the multiplication map. We have $$ (M_1+ N_1\sqrt{a} ) \cdot (M_2+ N_2\sqrt{a} ) = (M_1 M_2 + a N_1 N_2) + (M_1 N_2 + N_1 M_2) \sqrt{a}$$ so we ca write multiplication as $$(M_1,N_1) \cdot (M_2, N_2) = ( (M_1 M_2 + a N_1 N_2) , (M_1 N_2 + N_1 M_2) ).$$
Then $\Gamma$ is discrete inside the group of pairs of $2\times 2$ real matrices that have inverses under this multiplication map, i.e. those such that $M + N \sqrt{a}$ and $M - N \sqrt{a}$ are both invertible. This is isomorphic to $GL(2,\mathbb R) \times GL(2,\mathbb R)$.
But this is not the minimal group inside which $\Gamma$ is discrete. To find that, we need to take the Zariski closure of $\Gamma$, i.e. for all polynomial relations satisfied by $(M ,N) \in \Gamma$, we look at only those $M, N \in M_2(\mathbb R)$. This defines an algebraic group $G$, which always contains $\Gamma$, and that's the group in which $\Gamma$ is arithmetic.
These include $\det (M + N \sqrt{a})=1$ and $\det (M - N \sqrt{a}) =1$, which cut the group down to $SL(2,\mathbb R) \times SL(2,\mathbb R)$, which if $\Gamma =SL(2,\mathcal O)$ is as far as we go, but for $\Gamma$ the unitary group, there is another polynomial relation:
$$ \tau( g^T) Ag = A$$
i.e.
$$ ( M^T- N^T \sqrt{a}) A ( M+ N \sqrt{a} ) = A $$
and
$$ ( M^T+ N^T \sqrt{a}) A ( M- N \sqrt{a} ) = A $$
These two relations cut you down to a smaller group. Whatever it is, it's clearly not $SL (2,\mathbb R) \times SL(2, \mathbb R)$, because these relations are not satisfied for every element in $SL(2, \mathbb R)$. In fact, it's $SL(2,\mathbb R)$, for the reason YCor gave: Essentially, either $M+N \sqrt{a}$ or $M- N \sqrt{a}$ is determined by the other one, so you only need one $2\times 2$ matrix to determine the element.
In summary, you always want to express your group with coordinates over the integers, and relations defined by polynomials (ideally over the integers, or else remember to write down all Galois conjugates of your polynomial relations). Then you consider the real solutions of the same set of polynomial equations, and that will give you the right algebraic group. Before tricky relations that involve the Galois group like the one appearing in the definition of the unitary group, this will split as a product over the real places of your number field, but these extra relations will relate the projections to different places, giving a smaller group.
Best Answer
Motivation in mathematics is always a tricky question, but I'd call attention to one name you've omitted from your list: Serre. It's definitely worthwhile to look at his paper Cohomologie des groupes discrets in: Prospects in mathematics (Proc. Sympos., Princeton Univ., Princeton, N.J., 1970), pp. 77–169. Ann. of Math. Studies, No. 70, Princeton Univ. Press, Princeton, N.J., 1971.
The study of discrete subgroups in real Lie groups, starting with the classical modular group, has been a natural meeting place for geometry, number theory, group theory. Analogous groups over nonarchimedean local fields have become prominent in such questions as the Congruence Subgroup Problem; but here the nature of discrete subgroups is much less obvious. Serre points out right away the difficulty one has when taking products of locally compact groups over a mixture of fields (as in the use of adeles in number theory). For example, when $p$ is a fixed prime and $S$ consists of the infinite prime together with $p$, the $S$-arithmetic group $\mathrm{SL}_2(\mathbb{Z}[1/p])$ fails to be discrete in $\mathrm{SL}_2(\mathbb{R})$ as well as in $\mathrm{SL}_2(\mathbb{Q}_p)$. But it is discrete in the direct product of these two locally compact groups.
By working in this generality, one is able to unify considerably the study of discrete subgroups of locally compact groups along with related geometry and discrete group cohomology. Here the Bruhat-Tits buildings come into play along with classical symmetric spaces, etc.