$\mathbb {Z}[x]/(x ^ { n+1})=\{f\in\mathbb {Z}[x]|x ^ { n+1}=0\}=\{f\in\mathbb {Z}[x]|degree (f)\leq n\}$
Let the ideal $(x^{n+1})=K$, the basic idea behind quotient groups is to consider equivalence classes of the equivalence relation $f~g$ iff $f-g\in K$, so to consider it we look at our ring, but with the elements in the ideal set to zero.
You are correct that it is a quotient, but not a quotient of groups. It is instead a quotient of a ring by an ideal. This is defined similarly as for groups/normal subgroups:
Let $R$ be a ring, and let $\mathfrak{a}$ be an ideal of $R$. That is, $\mathfrak{a}$ is closed under addition/subtraction, and for all $r \in R, a \in \mathfrak{a}$, we have that $ra \in \mathfrak{a}$.
In such a case, we can define the quotient ring to be $$ R/\mathfrak{a} = \{r + \mathfrak{a}\mid r \in R\} $$ with multiplication and addition defined as you would hope them to be: $$ (r_1 + \mathfrak{a})(r_2 + \mathfrak{a}) = r_1r_2 + \mathfrak{a} $$ for example, and with zero element $0 + \mathfrak{a}$.
Anyhow, in the case of a principle ideal (i.e. one generated by a single element, $r$), we often write $\mathfrak{a} = (r)$. So this quotient above is just the quotient of the ring $\mathbb{Z}[x]$ by the ideal generated by $x^{n+1}$. That is, it kills all polynomial terms of degree higher than $n$.
