How to workout what elements of a quotient ring look like? I am trying to understand quotient rings. Firstly:
$$\frac{\Bbb Z[x]}{\langle x-1\rangle}$$
The above I can understand in a fairly naive way. Since the ideal is generated by a degree one polynomial, 1) we know by the division algorithm that we obtain a $0$-degree polynomial, and hence the above is isomorphic to the integers. Or 2) since $(x-1)\equiv 0$, $x\equiv 1$ and hence it is isomorphic to the integers. I am a little worried about the logic above, but I am fairly sure the result is correct.

Above is context, now for the question. I.e. TL:DR if you wish for context above.

$$\frac{\Bbb Z[x]}{\langle(x-1)(x-2)\rangle}$$
The ideal above is generated by a degree two polynomial, so by the division algorithm, we obtain all polynomials of the form: $a+bx|a,b\in\Bbb Z$. 
Now I don't know what information we have left to ponder over, something like the following? $x\equiv 1\pmod {x^2-3x+2}$ and $x\equiv 2 \pmod {x^2-3x+2}$
What more can we deduce?
 A: Yes, the elements of $$R := \Bbb Z [x] / \langle (x - 1) (x - 2) \rangle$$ "look like" $a + bx$, $a, b \in \Bbb Z$. Put more precisely, each element of $R$ has a unique representative in $\Bbb Z$ of degree $\leq 1$, like you say exactly because of the division algorithm.
This identification alone, however, does not determine the ring structure. The additive group underlying $R$ is isomorphic in the obvious way to the group $$\{a + bx : a, b \in \Bbb Z\}$$ under polynomial addition, which in turn is isomorphic to $\Bbb Z \oplus \Bbb Z$. The multiplicative structure is more complicated: If we denote the equivalence class in $R$ containing an element $p \in \Bbb Z[x]$ by $[p]$, we have
$$[a + bx] [c + dx] = [ac + (ad + bc) x + bd x^2],$$ and we can identify the linear representative of the latter class:
\begin{align}
[ac + (ad + bc) x + bd x^2]
&= [ac + (ad + bc) x + bd x^2 - bd(x - 1)(x - 2)]\\
&= [(ad + bc + 3bd) x + (ac - 2bd)].
\end{align}
A: From a computational point of view it amounts to saying that, each time you meet $x^2$, you can replace it with $3x-2$, $x^3$ will be replaced with $\,3x^2-2x=7x-6$, &c.
