If $$(\mathbb{Z}/\mathbb{Z}4) = \{\bar{0}, \bar{1}, \bar{2}, \bar{3}\}$$
How can it be possible that $x^2+2 \in (\mathbb{Z}/\mathbb{Z}4)[x]$?
In other words, how can the 2 value in $x^2+2$ be expressed as a member of ($\mathbb{Z}/\mathbb{Z}4)[x]$ if each element in ($\mathbb{Z}/\mathbb{Z4})[x]$ is a set?
$x^2+2$ here actually means $\bar 1\cdot x^2+\bar 2$.
I know it is super uncomfortable to accept the fact (fate) that sets can be used as elements of a group. But in the case of $\mathbb{Z}/4\mathbb{Z}$, indeed $sets$ are used as elements.
If you still feel very uncomfortable about this, try to use {a,b,c,d} to substitute $\{\bar 0,\bar 1,\bar 2,\bar 3\}$ with same operations. Then the polynomial above can be written as $bx^2+ax+c$, and we can even perform $a+b=b$ and $b+b+b+b=a$.
Just treat them as notations so that you will feel more comfortable.
The coefficients are $1$ and $2$... elements of ${\mathbb Z}_4$...
$1$ is the coefficient of $x^2$; $2$ the constant coefficient ... For each element (set) of $\mathbb Z/\mathbb Z4=\mathbb Z_4$, one can use any representative of that set, or residue class...
Thus $1=5=9=13=17...$ etc... all represent the same element...
Similarly, $2=6=10=14=18=\dots$ and so on.
There are $4$ such infinite sets, or elements, in $\mathbb Z_4$...
So, for instance, $5x^2+6$ is, or represents, the same polynomial here...