Let $n > 1$ be an integer. How to prove an algebra isomorphism $$K[\mathbb{Z}/n\mathbb{Z}] \cong K[t]/(t^n-1)?$$
Take $$K[\mathbb{Z}/n\mathbb{Z}] = K[\overline{0}, \overline{1}, \ldots, \overline{n-1}] = \{\sum_{i=0}^{n-1} k_i \overline{i}\mid k_i \in K\},$$ and $$K[t]/(t^n-1) = \{ \sum_{i=0}^{n-1} k_i t^i + (t^n-1) \mid k_i \in K\}.$$
I was trying to the case of $n=2$, $$\phi: K[\overline{0}, \overline{1}] \cong K[t]/(t^2-1).$$
If $\phi(\overline{1}) = t + (t^2-1)$, then $\phi(\overline{1}\cdot \overline{1}) = t^2 + (t^2-1) = 1 + (t^2-1) \neq \phi(\overline{1})$. So $\phi(\overline{1}) = \frac{1+t}{2} + (t^2-1)$.
Try defining an isomorphism that sends $\overline{a}$ to $t^a+(t^n-1)$. Remember that the group operation in $\mathbb{Z}/n\mathbb{Z}$ is what is normally written as addition, not multiplication. This means that in $K[\mathbb{Z}/n\mathbb{Z}]$, $\overline{a}\cdot\overline{b}=\overline{(a+b)}$, not $\overline{(a\cdot b)}$. In your example with $n=2$ that gives $\overline{1}\cdot\overline{1}=\overline{0}$, which is compatible with defining $\phi(\overline{a})=t^a+(t^2-1)$. Of course, you still must do more work to check that you really can define an isomorphism that way and that it works for any $n$.