"Subgroups" of a group correspond to "left coideal subalgebras" of a Hopf algebra. Why "subgroups" do not corresponds to "Hopf subalgebras" but "left coideal subalgebras"?
One way to think of this is to wonder how you're going to get well-behaved Hopf algebra quotients without being a left coideal, and then noting you don't need a stronger comultiplication condition than that, and in fact require it. So imposing "subcoalgebra" as well is a very strong restriction that eliminates a lot of perfectly nice quotients.
Morphism kernels are also properly interpreted as coinvariants. These are naturally expressible as normal coideal subalgebras, and are only Hopf subalgebras in very special cases (which can imply certain things regarding the Hopf algebra's structure and properties, which are again hardly necessary in general). You don't need stronger conditions than this to ensure that you have a morphism of Hopf algebras, and again eliminating perfectly nice morphisms (and quotients) tends to just make everything harder.
The downsides are notable. It becomes rather non-trivial to define a notion of short exact sequence, for example. While you can map a normal Hopf subalgebra of $H$ to a normal coideal subalgebra via $K\mapsto HK^+$, this is in general not an injective map. So naive attempts at defining short exact sequences run into the issue that they're not particularly well-defined, as the kernels aren't solely dependent on the Hopf subalgebra you'd like to use. A rather less naive approach becomes necessary.
As it turns out, this map turns out to be bijective when considering group algebras (and a few other broad classes of well-behaved Hopf algebras), and this is the (or at least "a") Hopf algebraic reason why group theory can focus entirely on the subgroups. You can also think of this as (partially) explaining why the theory of Hopf algebras isn't just a cut-and-paste job from group theory.