Let $\phi:\mathbb{Z}/12\mathbb{Z}\to\mathbb{Z}/3\mathbb{Z}$ be the homomorphism such that $\phi(1)=2$. Find the correspondence between $\mathbb{Z}_{12}/K$ and $\mathbb{Z}_3$ described by the first Isomorphism Theorem, where $\mathbb{Z}_3:=\mathbb{Z}/3\mathbb{Z}$, $\mathbb{Z}_{12}:=\mathbb{Z}/12\mathbb{Z}$, and $K=ker(\phi)$.
I know that $ker(\phi)=\{0,3,6,9\}$ and the cosets of $\mathbb{Z}_{12}$ are $\begin{matrix} K+0=\{0,3,6,9\} \\ K+1=\{1,4,7,10\} \\ K+2=\{2,5,8,11\} \end{matrix}$, and I know that the answer is supposed to be $$\mu(x)=\begin{cases} 0 & x=0+K \\ 2 & x=1+K \\ 1 & x=2+K \end{cases},$$ But I can't figure out why, even after looking at the first Isomorphism Theorem.
Hint: For any $a \in \{0, \ldots, 11\}$, notice that: $$ \phi(a) = \phi(\underbrace{1 + \cdots + 1}_{a \text{ times}}) = \underbrace{\phi(1) + \cdots + \phi(1)}_{a \text{ times}} = \underbrace{2 + \cdots + 2}_{a \text{ times}} = (2a \bmod 3) $$