The warm-up for any Hopf algebra construction is to try it on the two Hopf algebras $\mathbb K G$ and $C(G,\mathbb K)$, the group ring and ring of functions respectively for a finite group $G$.

In the group ring $\mathbb K G$, the group conjugation manifests in the forward direction. Recall that the structure maps on $\mathbb K G$ are simply the linearizations of the structure maps on $G$, where the comultiplication $\Delta$ is the duplication map $g \mapsto g\otimes g$ for $g\in G$ and the antipode $S$ is $g \mapsto g^{-1}$. Then the conjugation is:
$$ g\otimes h \mapsto S(g_{(1)}) \, h \, g_{(2)} $$
where I have used the Sweedler notation to denote the comultiplication, and the multiplication is simply concatenation. So this is a map $\mathbb K G^{\otimes 2} \to \mathbb K G$ extending $(g,h) \mapsto g^{-1}hg$. Incidentally, it's probably better to use the other conjugation $(g,h) \mapsto ghg^{-1}$, as that gives a left-action of $G$ on $G$, and you are writing the actor on the left. But I'll keep the ordering you gave in the question, for now. At the end, I'll come back to this, and it will be clear where the confusion is.

In $C(G,\mathbb K)$, everything is naturally reversed. In particular, for $G$ finite, $C(G,\mathbb K)$ has a basis given by the delta functions $\delta_g$ for $g\in G$. The multiplication is commutative and the comultiplication is anticommutative. We want the map $\delta_{g^{-1}hg} \mapsto \delta_g \times \delta_h$, or perhaps in the other order, depending on your conventions.

So let's read backwards. We first blow up $a = \delta_{g^{-1}hg}$ to $a_{(1)} \otimes a_{(2)} \otimes a_{(3)}$, where we think of $a_{(1)}$ as $\delta_{g^{-1}}$, etc. (of course, it isn't just that, but we'll pick out that term). Now we need to move the last term past the middle term, and antipode the first term: $S(a_{(1)}) \otimes a_{(3)} \otimes a_{(2)}$. Then the final multiplication makes sure that $S(a_{(1)})$ and $a_{(3)}$ are supported at the same points in $G$. So, all together, I think it's reasonable to call:
$$ a \mapsto S(a_{(1)})a_{(3)} \otimes a_{(2)} $$
a "coadjoint coaction", with the caveat as above that it's probably on the wrong side.

In any case, up to left and right, this is your second proposal. And left and right is hard, for the following reason. There seems to be no consensus in the quantum groups literature for whether the dual to $A \otimes B$ is $A^* \otimes B^*$ or $B^* \otimes A^*$. Or rather, it's clear from the representation theory of Hopf algebras that it must be the latter, but many of the early (and later) texts on Hopf algebras use the former dual when working in vector spaces (defining duals to Hopf algebras, etc.).

But also, for $C(G,\mathbb K)$ it doesn't matter: $ab = ba$. So really what you must do is check that your proposal is really a coaction, because in Hopf land this will pick out the difference.

In any case, I think you'll find when you work it out that this is not a coaction, exacly because $(g,h) \mapsto g^{-1}hg$ is no an action of groups (the $g$ is on the wrong side). If you do the correctly-sided action $(g,h) \mapsto ghg^{-1}$, then in $\mathbb K G$ this is $g\otimes h \mapsto g_{(1)}hS(g_{(2)})$, and in $C(G,\mathbb K)$ it is $a \mapsto a_{(1)}S(a_{(3)}) \otimes a_{(2)}$ (or the one you initially start with, if you have the opposite left-right convention when dualizing). Then you can just check directly that this is in fact a coaction of Hopf algebras.