One can avoid these set-theoretic issues either with grothendieck universes, or by looking at $\kappa$-small spaces for some cardinal: Fix a set $S$ of cardinality $\kappa$, and then just consider the set of all topologies on subsets of $S$. Every space of cardinality at most $\kappa$ is then homeomorphic to one of those, and they form a set. I believe the OP would be satisfied with an interesting topology on that set instead of "all topological spaces".

Ah. Then he probably deduces the unstable statement from the stable one? Once you know that $H^*(MO)$ is free (which you can prove less explicitly using the Milnor-Moore theorem), the unstable statement follows simply because the cohomology of $MO(r)$ and $MO$ agree in the necessary range of degrees.

Writing down such a homomorphism just corresponds to choosing a family of elements of degree $n_i$ in the target. Checking that it is an isomorphism in degrees $\leq 2r$ amounts to checking that certain terms are linearly independent and generate. This of course involves serious computations, the details should be explained by Stong.

For what it's worth, I've just convinced myself that it can't be done with 2 states, each of the 16 2-state rules has very simple behavior.

Do you mean Whitehead's problem? Isn't that about general abelian groups with vanishing Ext? What exactly do you mean by "there are whitehead groups whose duals are products"? (I'm confused since as far as I understand Whitehead's conjecture IS settled, and it's known that the existence of non-free whitehead groups is independent of ZFC)

Yes, since for any $m$ $a\uparrow\uparrow n$ converges to a fixed value mod $m$, $a\uparrow\uparrow \infty$ gives a well-defined element of the profinite completion $\widehat{\mathbb{Z}}$.

(but I agree with Denis T's comment in that if you're looking for intuition, it should be "flatness is a rare and special phenomenon")