Understanding the ideal generated by a polynomial

Put in simple words,

In any ring $R$ (with unit!), the principle ideal $(a)$ generated by $a\in R$ consists of nothing but the multiples of $a$.

I give here just a heuristic:

I like to think of the quotient ring $R/(a)$ as having the same elements of $R$ but the equality is modified so that all elements of $(a)$ be equal to zero in the quotient. Observe that it is enough to require $a=0$, because $ras=0$ and $\sum r_ias_i=0$ already follows by the properties of equality.

In the case of polynomial ring, a factor like $R[t]/(f)$ with, say $f=t^n-a_{n-1}t^{n-1}-\dots-a_1t-a_0$ will always be represented by the set of polynomials of degree $<n$, because in the quotient ring we have $$t^n=a_{n-1}t^{n-1}+\dots+a_1t+a_0$$ So that, any time a $t^n$ appears, it can be replaced by a degree $<n$ polynomial in $R[t]/(f)$.

When the ring is commutative, as in your case, things are way simpler and easier:

$$\forall\,a\in R\;,\;\;(a):=\{ra\;;\;r\in R\}$$

And thus in your very particular case, the ideal generated by a polynomial $\;f(x)\in R[x]\;$ is just the set of all products of $\;f\;$ by elements of $\;R[x]\;$ .

A two-sided ideal $I$ of a ring $R$ is naturally the kernel of the surjective homomorphism $$ R \twoheadrightarrow R/I. $$

Take your example, where $R = \Bbb{Z}/2\Bbb{Z}[t]$ and $I = (f) = (t^2 + t + 1)$. In the quotient (the image of the map), the polynomial $f$ is identified with $0$. But if $f = 0$, then $af = a0 = 0$ and $fb = 0b = 0$ for any $a, b \in R$. The fact that the ideal must be closed under multiplication by any element in the ring on either side is forced by the desire for the ideal to be a kernel.

Since $t^2 + t + 1 = 0$ in $R/I$, any time that you see $t^2$, you can replace it by $-(t + 1)$ or just $t + 1$, as the ground field is $\Bbb{Z}/2\Bbb{Z}$.

Try working out the sum $(at + b) + (ct + d)$ and product $(at + b)(ct + d)$ of two generic elements in $R/I$ to get a sense of the structure of this ring. Hint: it's not very big.