$S$ is a subring of $R$, $J$ is an ideal of $R$, and $\Phi: (S + J) \to S/(S \cap J)$ where $\Phi(a + b) = a + (S \cap J)$. We need to show that $\Phi$ is well defined.
Suppose that $a_1 + b_1 = a_2 + b_2$ for some $a_1, a_2 \in S$ and $b_1, b_2 \in J$.
Step 1: $a_1 + b_1 = a_2 + b_2 \Rightarrow$
Step 2: $a_1 - a_2 = b_2 - b_1 \in (S \cap J)$
Step 3: $⇒ a_1 + (S \cap J) = a_2 + (S \cap J)$.
Which we wanted to show.
But my question is how we went from Step 2 to Step 3? How Step 2 implies Step 3?
I've tried it a lot but's it's still not clicking in my head.
Thanks
Step 2 shows that $a_1$ and $a_2$ differ by an element of the ideal $S\cap J$, and we can write $a_1=a_2+(b_1-b_2)$ and $a_2=a_1+(b_2-b_1)$.
One way to think of an ideal is as a cluster of elements of a ring that we would like to imagine to behave collectively as $0$. This perspective motivates the axioms for an ideal: an ideal must absorb products ($x\cdot I=I$) because the product of anything with $0$ is still $0$ and it must be normal as a subgroup (i.e. $x+I=I+x$) because $0$ commutes with everything under addition (as well as following the other group axioms). Notice how these axioms need not be obeyed pointwise by elements of the ideal. We can have things like $x\cdot i_1=i_2$ (with $i_1,i_2\in I$), showing that the particular element $i_1\in I$ doesn't absorb this product by itself; it is only important that $I$ behaves like $0$ as a whole and that $x\cdot I=I$.
With this in mind, we can develop the idea of a coset $x+I$ of an ideal. If $I$ behaves as a cluster of elements behaving as $0$, adding these elements from $I$ to $x$ should not change the value of $x$. In other words, if all elements of $I$ are grouped together as $0$, this necessitates that all elements of $x+I$ should be thought of as equal, because adding $0$ to $x$ shouldn't change its value.
Finally, we can apply this intuition to your problem. If we think of $S\cap J$ as behaving collectively as a $0$ element, the fact that $a_1=a_2+(b_1-b_2)$ should let us think that we can attain $a_2$ from $a_1$ by adding an instance of $0$ (in this case $b_1-b_2$), so we ought to think of them as equal. The formal translation of this is that they have the same coset with respect to $S\cap J$. The algebra relating to cosets justifies this perspective: since $a_1=a_2+(b_1-b_2)$, we can write $$ a_1+S\cap J=[a_2+(b_1-b_2)]+S\cap J=a_2+[(b_1-b_2)+S\cap J]=a_2+S\cap J, $$ relying on the (group-theoretic) fact that adding $(b_1-b_2)$ to $S\cap J$ shuffles it but results in the same set $S\cap J$.