Given two spectra $E$ and $F$, how might we get a handle on the contents of the spectrum $E \wedge F$? One thing we could try is to produce an interesting map involving $E \wedge F$ as its source or target that relates to things we already understand. Let's assume complex orientations for $E$ and $F$, so maps $u: MU \to E$ and $v: MU \to F$, which we combine to get a map $MU \wedge MU \to E \wedge F$. The homotopy of the spectrum $MU \wedge MU$ carries the universal example of a formal group law isomorphism, and so the map we constructed selects an isomorphism of the formal group laws associated to the composite orientations $MU \xrightarrow{u} E \xrightarrow{\eta_F} E \wedge F$ and $MU \xrightarrow{v} F \xrightarrow{\eta_E} E \wedge F$. In some cases we are lucky enough to produce an isomorphism of the formal groups of $E$ and of $F$, like with your computation of interest when $E = F = H\mathbb{F}_2$: $$\operatorname{Spec} \pi_* H\mathbb{F}_2 \wedge H\mathbb{F}_2 = \operatorname{Aut}(\hat{\mathbb{G}}_a).$$ This is definitely not going to happen in general, since the homotopy of the smash product can be either too complicated or not complicated enough, or the composite orientations might not compare well with the originals --- they might be "damaged" in some way through pushforward. In the case of the dual Steenrod algebra it is in some sense a statement about spread-out-ness of the objects involved. But, as far as I know, this is as deep an explanation as you can get at present.

I'm told that something similar happens in the odd primary case, but involves automorphisms of the formal additive "supergroup". I have no idea how it works, though, and is probably generally related to my poor understanding of graded commutativity and odd dimensional phenomena in general. Definitely it is mentioned in brief in COCTALOS; searching on 'super' will bring it right up if you want to read a few sentences more.

Given two spectra $E$ and $F$, how might we get a handle on the contents of the spectrum $E \wedge F$? One thing we could try is to produce an interesting map involving $E \wedge F$ as its source or target that relates to things we already understand. Let's assume complex orientations for $E$ and $F$, so maps $u: MU \to E$ and $v: MU \to F$, which we combine to get a map $MU \wedge MU \to E \wedge F$. The homotopy of the spectrum $MU \wedge MU$ carries the universal example of a formal group law isomorphism, and so the map we constructed selects an isomorphism of the formal group laws associated to the composite orientations $MU \xrightarrow{u} E \xrightarrow{\eta_F} E \wedge F$ and $MU \xrightarrow{v} F \xrightarrow{\eta_E} E \wedge F$. In some cases we are lucky enough to produce an isomorphism of the formal groups of $E$ and of $F$, like with your computation of interest when $E = F = H\mathbb{F}_2$: $$\operatorname{Spec} \pi_* H\mathbb{F}_2 \wedge H\mathbb{F}_2 = \operatorname{Aut}(\hat{\mathbb{G}}_a).$$ This is definitely not going to happen in general, since the homotopy of the smash product can be either too complicated or not complicated enough, or the composite orientations might not compare well with the originals --- they might be "damaged" in some way through pushforward. In the case of the dual Steenrod algebra it is in some sense a statement about spread-out-ness of the objects involved. But, as far as I know, this is as deep an explanation as you can get at present.

I'm told that something similar happens in the odd primary case, but involves automorphisms of the formal additive "supergroup". I have no idea how it works, though, and is probably generally related to my poor understanding of graded commutativity and odd dimensional phenomena in general. Definitely it is mentioned in brief in COCTALOS; searching on 'super' will bring it right up if you want to read a few sentences more.

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Your mentioning of $\mathbb{R}\mathrm{P}^\infty$ is somewhat separate. The reason the Steenrod operations show up there is that $\mathbb{R}\mathrm{P}^\infty$ is a $B\mathbb{Z}/2$ and $\mathbb{Z}/2 = \Sigma_2$ is a symmetric group permuting the inputs to a cup product. It's an altogether different miracle that this recipe for constructing cohomology operations exhausts all of $H\mathbb{F}_2$'s.