What I'd like to say is, to some degree, commentary and expansion on what some people have said in the comments above.
I feel that the answer to (Q1), about whether there is a "high-concept" explanation for the dual Steenrod algebra, is "no" at the current point in time. Even further, I feel like the attempt to do so right now would be misleading. We can assign some high-concept descriptions to phenomena in terms of formal group laws, but trying to give a high-concept reason why the Steenrod algebra takes the form it does would be misinterpreting the role of the Steenrod algebra.
We construct the stable homotopy category from the category of topological spaces. Based on its construction, we can say a lot of things in older and more modern language. It's a tensor triangulated category; it's the homotopy category of a symmetric monoidal stable model category; it's the homotopy category of a stable $\infty$-category; it's some kind of "universal" way to construct a stable category out of the category of topological spaces. Based on this, we can describe a lot of the foundational properties of the stable homotopy category and the category of spectra. However, there are a ton of categories that satisfy basically the same properties. Without doing more work, we don't understand anything about specifics that distinguish the stable homotopy category from any other example.
These kinds of computations originate with the Hurewicz theorem, the Freudenthal suspension theorem, Serre's computation of the cohomology of Eilenberg-Mac Lane spaces and his method for computing homotopy groups iteratively, and Adams' method of stepping away from Postnikov towers and upgrading Serre's method into the Adams spectral sequence. Before these, you would have no reason to necessarily believe that the stable homotopy category isn't equivalent to, say, the derived category of chain complexes over $\mathbb{Z}$ or some other weird differential graded algebra. Before you have these, you don't have Milnor's computation of the homotopy groups of $MU$ or $MU \wedge MU$, and you don't have Quillen's interpretation in terms of the Lazard ring.
In short, understanding the Steenrod algebra and its dual are prerequisites for all the qualitative things we understand about the stable homotopy category to distinguish it from another example. Right now there is not a door to the stable homotopy category that comes from formal-group data, as much as we would like one. As a result, I guess I feel like trying to assign a high-concept description to the Steenrod algebra is backwards right now.
The fact that often computations come first and the conceptual interpretations second is something that gives the subject some of its flavor. Are there high-concept explanations why vector bundles should have Stiefel-Whitney classes? Why complex vector bundles should have Chern classes? Why the "quadratic" power operations in mod-$2$ cohomology should generate all cohomology operations, and why all the relations are determined by those coming from composing two?
This is definitely not to say that such a description wouldn't be desirable. It would be very desirable to have a more direct route from concepts to the stable homotopy category, because constructing objects that realize conceptual descriptions can be very difficult.
Consider $G$ of order $2$ acting trivially on the sphere spectrum $S^0$. In this key example, smashing with $H\mathbb Z/2$ drastically fails to commute with $^{hG}$: If it did commute, then the mod $2$ homology of the homotopy fixed point spectrum would be the homotopy of $(H\mathbb Z/2)^{hG}$, so $\mathbb Z/2$ in nonpositive dimensions and $0$ in positive dimensions. But the (solved) Segal conjecture shows that the homotopy fixed point spectrum is connective.
Best Answer
Propositions 17.10 and 17.11 of Switzer's book seem to be examples of things working nicely.
For those who don't have the book to hand: let $E$ be a commutative ring spectrum such that $E_\ast(E)$ is flat as a right module over $E_\ast=E_\ast(*)$, and let $X$ be any spectrum.
Let $\psi_X\colon E_\ast(X)\to E_\ast(E)\otimes_{E_\ast} E_\ast(X)$ be the coaction map, and let $c\colon E_\ast(E)\to E_\ast(E)$ be the conjugation map (which switches the left and right counits).
Then for any elements $a\in E^\ast(E)$, $y \in E^\ast(X)$ and $u\in E_\ast (X)$ with $\psi_X(u) = \sum_i e_i\otimes u_i$, we have $$ a\cdot u = \sum_i \langle a,ce_i\rangle u_i $$ and $$ \langle ay,u\rangle = \sum_i (-1)^{|y||e_i|}\langle a,e_i\langle y,u_i\rangle \rangle, $$ where $\langle \cdot , \cdot \rangle\colon E^\ast(X)\otimes_{E_\ast} E_\ast(X)\to E_\ast$ is the Kronecker pairing.