Meaning of the antipode in Hopf algebras?

$\bullet$ Let $(C, \Delta, \varepsilon)$ a $k$-coalgebra, and $(Α, \mu, \eta)$ a $k$-algebra. Inside the $k$-v.s. of linear maps from $C$ to $Α$, i.e. inside: $Hom_{k}(C,A)$, we define a bilinear multiplication between $f, g \in Hom_{k}(C,A)$ through: $$ C \stackrel{\Delta}{\longrightarrow} C \otimes C \stackrel{f \otimes g}{\longrightarrow} A \otimes A \stackrel{\mu}{\longrightarrow} A $$ In other words, for any $f,g \in Hom_{k}(C,A)$, we will have: $$f \star g = \mu \circ (f \otimes g) \circ \Delta$$ or equivalently in Sweedler's notation: $$ (f \star g)(c) = \sum f(c_{(1)})g(c_{(2)}) $$ for any $c \in C$.

The multiplication $\star \;$ defined above, is clearly bilinear, and we can easily see that $f \star g \in Hom_{k}(C,A)$. The multiplication $\star \;$, will be called convolution product.

Now, it is relatively easy to show, that, the $k$-v.s. $Hom_{k}(C,A)$ equipped with the convolution product becomes a $k$-algebra. Associativity can be shown via a straightforward computation, while the unit element will be $\eta\circ\varepsilon \in Hom_{k}(C,A)$.

$\bullet$ Let us now pass to a concrete example of the above construction: Let $Η$ a $k$-bialgebra. If we denote by $H^{c}$ the underlying $k$-coalgebra and with $Η^{a}$ the underlying $k$-algebra. Inside the $k$-v.s. $Hom_{k}(H^{c}, Η^{a})$ i.e. inside $End_{k}(H)$, we consider the convolution product and the algebra structure defined above. We note that the identity map $Id_{H} : H \rightarrow H$ is an element of this algebra.

Definition: Let $Η$ be a $k$-bialgebra. A linear map $S: H \rightarrow H$ will be called antipode of the $k$-bialgebra $Η$, if $S$ is the inverse of the identity map $Id_{H} : H \rightarrow H$ with respect to the convolution product. In other words, if: $$ S \star Id_{H} = Id_{H} \star S = \eta \! \circ \! \varepsilon $$ or equivalently, if $$ \mu \circ (S \otimes Id) \circ \Delta = \mu \circ (Id \otimes S) \circ \Delta = \eta \! \circ \! \varepsilon $$ or equivalently (using Sweedler's notation for the comultiplication): $$ \sum S(h_{(1)})h_{(2)} = \sum h_{(1)}S(h_{(2)}) = \varepsilon(h)1_{H} $$ for any $h \in H$.

Note that, according to the above definition, a given bialgebra does not necessarily have an antipode. However, if it has one, it is easy to show that it will be unique (the proof is similar with the one for the uniqueness of the inverse element in a group).

Intuitively, you can say that the antipode map is a kind of generalization of the inverse element in a group. This intuitive interpretation is supported by the fact that for group Hopf algebras -which have been among the first and most well-studied examples of Hopf algebras- the antipode map is given by: $S(g)=g^{-1}$.

The importance of the antipode, from the viewpoint of representation theory, has to do with the fact that the presence of an antipode (i.e. the Hopf algebraic structure), allows one to construct dual representations in a way which is reminiscent of the group representation theory. You can see here, Ch. 9, par. 9.3, p.442, for more details.

P.S.: In the last part of your post, you say

"...looks like without that axiom the product and co-product would be unrelated to each other"

Note that this is not correct. The above relation (in the antipode's definition) is certainly a "constraint" between the algebra and coalgebra structures. However, this is not the only "constraint" between the multiplication and the comultiplication. They need to satisfy a certain constraint for the bialgebra structure (see here for details) and another one (i.e. the present one, described above in the antipode's definition) for the Hopf algebra structure.

A Hopf algebra is an example of an object called a Hopf monoid, which you can define in any symmetric monoidal category. If you have a look at the definition, it looks almost exactly like the definition of a group object, but with tensor products instead of the usual products, and with compatible comultiplication.

However, in any category with products, every object automatically comes with a comultiplication given by the diagonal map, and every monoid object is canonically a bimonoid. Thus, if you consider the category of sets with the Cartesian product as the monoidal product, then a Hopf monoid in the category of sets is precisely a group, and the antipode map is just the inverse map $g \mapsto g^{-1}$ of the group! Similarly Hopf monoids in the category of smooth manifolds are precisely Lie groups.

In the category of $R$-modules ($R$ commutative) with the usual tensor product, a Hopf monoid is just a Hopf algebra. So Hopf algebras are really like "groups" in the category of $R$-modules, and the antipode should be thought of as being like the group inverse.

From the point of view of representation theory, the antipode is essential because it allows you to form the dual representation (just as the antipode=inversion for group representations). This is why, for instance, Lie algebras have dual reps: the universal enveloping algebra is a Hopf algebra!

As an aside: thinking about groups as Hopf monoids in the category of sets is very convenient, because it immediately follows that the tensor algebra $T(V)$ of an $R$-module $V$ is a Hopf algebra, and so is the group ring $R[G]$ of a group $G$.