In Majid's quantum group primer at the beginning of Chapter 3, page 18, he's proving that if $H'$ and $H$ are dually paired bialgebras or Hopf algebras, the coadjoint action $$ \operatorname{Ad}^*_\phi(h)=\sum h_2\langle\phi,(Sh_1)h_3\rangle $$ respects the coproduct by writing (in Sweedler notation, with summation signs dropped) $$ \begin{align*} \operatorname{Ad}^*_{\phi_1}(h_1)\otimes\operatorname{Ad}^*_{\phi_2}(h_2)&=h_2\langle\phi_1,(Sh_1)h_3\rangle\otimes h_5\langle\phi_2,(Sh_4)h_6\rangle\\ &= h_2\otimes h_5\langle\phi,(Sh_1)h_3(Sh_4)h_6\rangle\\ &= h_2\otimes h_3\langle\phi,(Sh_1)h_4\rangle=\Delta\circ\operatorname{Ad}^*_\phi(h) \end{align*} $$
Here $S$ is the antipode, but I can't follow how the antipode axioms imply the second to last equality $$ h_2\otimes h_5\langle\phi,(Sh_1)h_3(Sh_4)h_6\rangle= h_2\otimes h_3\langle\phi,(Sh_1)h_4\rangle. $$
The antipode identities say for all $h\in H$ that $$S(h_1)h_2 = h_1 S(h_2)=\epsilon(h)1_H.$$
More formally, $$m\circ (S\otimes \operatorname{id}) \circ\Delta = m\circ(\operatorname{id}\otimes S)\circ\Delta = \eta\circ\epsilon,$$ where $\eta\colon k\to H$ is the unit.
As a consequence, you also have identities like $$h_1\otimes h_2\otimes S(h_3)h_4=h_1\otimes h_2\otimes h_3 S(h_4) = h_1\otimes h_2 \epsilon(h_3)\otimes 1 = h_1\otimes h_2\otimes 1,$$ where we applied the above identity in the third tensorand.
The indexed items are still themselves elements of the Hopf algebra, so the identity applies to them as well: you just have to perform the manipulation within the context of the rest of the indices. For example, saying $h_3 S(h_4)=\epsilon(h_3)$ all by itself is wrong if these are Sweedler notation indices. Or more accurately, it's "not even wrong", because it's invalid: for Sweedler indices of 3 and 4 to make any sense in an expression there must be indices for 1 and 2, as well.
This may be more clear if we unmask the coassociativity that allows Sweedler notation a bit: $$h_1\otimes h_2\otimes h_3 S(h_4) = h_1\otimes h_2\otimes (h_3)_1 S((h_3)_2).$$