Fundamental theorem of isomorphism says that if $\phi : G \to \bar G$ is a group homomorphism, then the mapping $\bar\phi : G/\ker(\phi) \to \phi(G)$, defined by $\bar\phi(g\ker(\phi)) = \phi(g)$, is an isomorphism. In other words, $G/\ker(\phi) \cong \phi(G)$
Now we know that there are exactly $4$ homomorphisms from $Z_{20}$ onto $Z_{10}$ : there must be isomorphism between them according to theorem. For that order of $\ker(\phi)$ must be $2$ . But since by the property of homomorphism, an identity element gets mapped with identity, so all the elements of $\ker(\phi)$ must be identity element of $Z_{20}$ and they must be two at the same time we know identity element is unique.
You ask how the kernel of a homomorphism can ever have more than one element.
Consider this example. Let $G$ and $H$ be groups. Define the map $$ \phi: G\to H $$ by $$ \phi(g) = e_H. $$ for all $g\in G$. That is, $\phi$ maps all elements in $G$ to the identity element in $H$. This is a homomorphism because $$ \phi(g)\phi(h) = e_he_h = e_h = \phi(gh) $$ for all $g,h\in G$. And for this map you have $$ \ker\phi = \{g\in G: \phi(G) = e_h\} = G. $$ Now if your homomorphism is injective (one-to-one) then indeed only the identity in $G$ maps to the identity in $H$. It is, in fact, true that a homomorphism is injective exactly when the kernel is trivial.
So, if you would more examples of homomorphisms that have non-trivial kernels, just look at examples of homomorphisms that are not injective.