If [tex]\phi[/tex] is a homomorphism, then we have, for every [tex]g,h \in G[/tex]
[tex]\phi(gh) = \phi(g)\phi(h)[/tex]
Since G is abelian, we have [tex]gh=hg[/tex], and thus [tex]\phi(gh)=\phi(hg)[/tex]
But we also have
[tex]\phi(gh)=\phi(g)\phi(h)=\phi(h)\phi(g)=\phi(hg)[/tex]
which proves that G' is abelian.
In other words, for every [tex]x,y \in G'[/tex], you have [tex]xy=yx[/tex], because there exist [tex]g,h \in G[/tex] such that [tex]x=\phi(g),\ y=\phi(h)[/tex], and you can think of [tex]xy[/tex] as [tex]\phi(g)\phi(h)[/tex], and of [tex]yx[/tex] and [tex]\phi(h)\phi(g)[/tex]
Then, you observe that [tex]\phi(g)\phi(h)=\phi(h)\phi(g)[/tex] because they mean [tex]\phi(gh)=\phi(hg)[/tex], and [tex]gh=hg[/tex] by hypothesis, because G is abelian.
