Hi Ryan,
Let me elaborate on my answer to your previous question. Somehow, deforming only the algebra structure is easy, in the sense that if you give some generators and some relations you're sure to get... an algebra. So just take the same generators as for $U(\mathfrak{sl}_2)$, some random relations whose quasi-classical limit gives you the defining relations of $U(\mathfrak{sl}_2)$ and you're done. Of course it's not a very interesting approach, but at least you are sure to get something matching your requirements.
On the other hand, deforming the Hopf structure is hard: constructing a well defined algebra map $\Delta$ is easy, but coassociativity is a hard condition which is of course not guaranted at all if you pick something at random.
First of all, it's strictly speaking not completely true that $K=q^H$ since $H \not\in U_q(\mathfrak{sl}_2)$, but it is still the right intuition, justified by the "formal" setting below. Then the same argument as in my previous answer apply:
$$\Delta(K)=q^{\Delta(H)}=q^{H\otimes 1+1\otimes H}=(q^H \otimes 1)(1 \otimes q^H)=K\otimes K$$
All of this is more natural if you take the formal version $U_{h}(\mathfrak{sl}_2)$ which is a topological $\mathbb{C}[[\hbar]]$-Hopf algebra. In that case $H$ is a true generator and there are many reasons for which you want to let the "Cartan part" really undeformed. It is, indeed, a general fact: for a semi simple Lie algebra $\mathfrak{g}$, the sub-Hopf algebra generated by the $H_i$'s is isomorphic as an Hopf algebra to $U(\mathfrak{h})[[\hbar]]$, and the isomorphism is just the identity $H_\mapsto H_i$. In fact, it is known that $U_{h}(\mathfrak{g})$ is isomorphic as an algebra to $U(\mathfrak{g})[[\hbar]]$, but now the isomorphism is far from being trivial. Still, it clearly show that the important things is the whole Hopf algebra structure.
Therefore we know the coproduct for $K$, or at least we expect that there exists an Hopf structure such that $\Delta(K)=K\otimes K$. Now you can do some computation, write down the constraints imposed by co-associativity, still trying to get something "familiar" (ie s.t. $E$ and $F$ are close from being primitive). I may be wrong but I'm not sure that there is another way of finding something explicit, so what you did was right.
Now for general $\mathfrak g$ you can try to "glue together" copies of $U_q(\mathfrak{sl}_2)$. Again there is nothing obvious here, and finding these formulas was a great achievement. It seems to me, for example, that no similar deformation for non semisimple Lie algebras are known except in a few cases.
As I said in the abovementionned answer, the very much existence of a non-trivial deformation of the Hopf structure is a highly non-trivial fact, which "could have been false". The deep reason for which such object exists is really related to the theory of Drinfeld associators.