Let $K(G)$ be the commutator subgroup. It is said that in the quotient space $G/K(G)$ $$\begin{align} [a][b]&=[a][b][b^{-1}a^{-1}ba]\\ &=[a][b][b^{-1}][a][b][a]\\ &=[b][a]. \end{align}$$
Where does the first equality come from?
I only see that since $ab=[a,b]ba$, I get $$\begin{align} [a][b]&=[ab]\\ &=[[a,b]ba]\\ &=[aba^{-1}b^{-1}ba]\\ &=[a][b][a^{-1}b^{-1}ba] \end{align}$$
You can't prove that the commutator subgroup, your $K(G),$ is abelian, because in general it isn't.
For instance, $K(S_n)=A_n$, which is not abelian ($n\gt3$).
What you have above is the proof that $G/K(G)$ is abelian. That's the quotient of $G$ by $K(G)$. It's called the abelianization.
Since it's the first equality that bothers you, just note that in this quotient, everything in the commutator subgroup is set equal to $\bar e.$ But the commutator subgroup contains each $b^{-1}a^{-1}ba.$