How to deal with polynomial quotient rings

To treat your special case $$\mathbb{Z}_{2}[X] / (x^4+1),$$ its elements are the (classes) of the remainders of Euclidean divisions by $x^4+1$, so $$\mathbb{Z}_{2}[X] / (x^4+1) = \{ [a_0 + a_1 x + a_2 x^2 + a_3 x^3] : a_i \in \mathbb{Z}_{2} \}.$$ (Here I am using brackets to denote residue classes.)

You sum the classes normally, and as to the product, you first take the product, and then take the remainder of Euclidean division by $x^4+1$. Alternatively, you take the product, and then use the relation $x^4 \equiv -1$ (which is the same as $x^4 \equiv 1$ in this case, as $1 = -1$ in $\mathbb{Z}_{2}$) to reduce the result. So it's not really different from computing in $\mathbb{Z}_{m}$, where you first take an ordinary sum or product, and then take the remainder modulo $m$.

So for instance $$ [1 + x^2] \cdot [1 + x^3] = [1 + x^2 + x^3 + x^5] = [1 + x + x^2 + x^3], $$ using one of the two methods above.

The general case of $F[x] / (f(x))$, with $F$ a field, is similar, you get the classes of the remainders of Euclidean division by $f(x)$.

Often you can use the universal properties (of quotients and polynomial algebras) in order to simplify the rings. For example,

$\mathbb{F}_2[x]/(x^4+1) = \mathbb{F}_2[x]/((x+1)^4) \cong \mathbb{F}_2[y]/(y^4)$

and this cannot be simplified anymore. You just "cut" the polynomials in $y$ above $y^4$. Of course $y^5$ is still there, but it equals zero. And $1+y$ is a unit, since $1+y+y^2+y^3$ is inverse to it (more generally, unit + nilpotent = unit, this is the geometric series).

Often the Chinese Remainder Theorem helps to simplify the rings. It shows that if $f \in K[x]$ decomposes as $f_1^{k_1} \cdot \dotsc \cdot f_n^{k_n}$ with irreducible $f_i$, then $K[x]/(f) \cong \prod_{i=1}^{n} K[x]/(f_i^{k_i})$.

For example, $\mathbb{R}[x]/(x^2-1) = \mathbb{R}[x]/((x+1)(x-1)) \cong \mathbb{R}[x]/(x+1) \times \mathbb{R}[x]/(x-1) \cong \mathbb{R} \times \mathbb{R}$. Another interesting example is $\mathbb{Q}[x]/(x^n-1) \cong \prod_{d|n} \mathbb{Q}(\zeta_d)$.

Since the CRT holds for arbitrary rings, we can also apply it to $\mathbb{Z}/m[x]$. But the problem is that this ring is not a PID anymore (when $m$ is not a prime), so that we have no nice factorizations and even when they exist the corresponding ideals are not coprime and therefore CRT doesn't apply. For example, I don't think that we can really simplify $\mathbb{Z}/4[x]/(2x)$. The elements have the form $a_0+a_1 x+ a_2 x^2 + \dotsc$ with integers $a_i$, where we can calculate $a_0 \bmod 4$ and $a_i \bmod 2$ for $i>0$.