why $(r+I)(s+I) = rs + I$ in the quotient ring $R ?$
Solution 1:
The multiplication in the quotient ring is not defined by $$ (a+I)(b+I)= \{\,(a+i_{1})(b+i_{2}): (i_{1},i_{2})\in I \times I\,\} $$ but by $$ (a+I)(b+I)=ab+I. $$ This is a definition, nothing else. Why do we define it in this way? Because it does what we want, together with $$ (a+I)+(b+I)=(a+b)+I, $$ namely it makes $A/I$ into a ring and \begin{align} \pi\colon A &\to A/I\\ a&\mapsto a+I \end{align} a ring homomorphism.
The only thing to show is that the definitions are “correct”: if $a_1+I=a_2+I$ and $b_1+I=b_2+I$, then we should have $$ a_1b_1+I=a_2b_2+I $$ that is $$ a_1b_1-a_2b_2\in I. $$ This is true because $$ a_1b_1-a_2b_2=a_1b_1-a_1b_2+a_1b_2-a_2b_2= a_1(b_1-b_2) + (a_1-a_2)b_2 $$ and, by hypothesis, $a_1-a_2\in I$ and $b_1-b_2\in I$; apply the properties of $I$ to end the proof. Similarly for the addition. The verification of the ring properties is easy.
Solution 2:
You have a ring $R$ and an ideal $I\subseteq R$ (note that is is not a member of $R$, it is a subset).
Now you define an equivalence relation by $\forall a,b \in R, a\sim b \iff a-b \in I$
Then you look at $R/\sim = \left\{\overline{a}\mid a \in R\right\}$ where $\forall a \in R,\overline{a}=\left\{b \in R \mid b \sim a\right\}=a+I$ is the equivalence class of $a$.
Not you want to define $\forall a,b \in R,\overline{a}+\overline{b}=\overline{a+b}$ and $\overline{a}\times\overline{b}=\overline{a\times b}$ but to be able to do that, you need $\forall a,b,c,d \in R,a\sim c \land b \sim d \implies a+b \sim c+d$ and the same property with $\times$. Because otherwise, you would have $\exists a,b,c,d \in R,\overline{a+b}=\overline{a}+\overline{b}=\overline{c}+\overline{d}=\overline{c+d}$ and $\overline{a+b}\not= \overline{c+d}$ which is absurd. You can think of it this way: You want to be able to define the sum and the product of two equivalence classes so that it is independent of the representatives you chose.
And the fact that $I$ is an ideal gives this equivalence property those property (compatibility with $+$ and $\times$).