Understanding the idea of the proof of "For two ideals $I$ and $J$ prove that prove that $I(R/J)=(I+J)/J$" I was reading this question, but I could not understand the proof of why for two ideals $I$ and $J$ we have $I(R/J)=(I+J)/J$.
Here is the proof given in this link:

Well, if you use the definition, it's obvious: an element of $I\cdot R/J$ is  the ideal of $R/J$ generated by $I$, so it is the set of finite sums of the form:
$$\Bigl\{\sum_k i_kr_k+J\bigm\vert i_k\in I, r_k\in R\Bigr\}.$$
Now, as $I$ is an ideal in $R$, the set of these finite sums u=is nothing but $I$ itself, whence the formula
$$I\cdot R/J=(I+J)/J.$$
Note that, by the second isomorphism theorem,  the latter quotient is isomorphic to $I/I\cap J$.

Here is what I know:
I know how to show that $IM$ is a submodule of $M$ and hence it is an ideal of $M.$
Here is what I do not understand in the solution above:
1- Why $I \cdot R/J$ must be generated by $I$?
2- Why &quot;as $I$ is an ideal in $R$, the set of these finite sums u=is nothing but $I$ itself&quot; and why that leads to that $I(R/J)=(I+J)/J$ are not we moding out by $J$ in the RHS of this formula?
Could anyone explain these points to me please?
Given commutative unital rings $R$ and $S$ such that there exists a ring homomorphism $\varphi : R \to S,$ we define the extension of an ideal $I$ of $R$ to be the ideal $$I^e = \varphi(I) S = {\left \{\sum_{k = 1}^n s_k \varphi(i_k) \,|\, n \geq 0 \text{ is an integer, } i_k \in I, \text{ and } s_k \in S \text{ for all } k \right \}}$$ of $S.$ For an ideal $J$ of $R,$ there is a ring homomorphism $\pi : R \to R/J$ defined by $\pi(r) = r + J.$ Putting this together with the above shows that $I^e = \pi(I)(R/J)$ consists of the elements $$\sum_{k = 1}^n (r_k + J)(i_k + J) = \sum_{k = 1}^n (r_k i_k + J) = {\left(\sum_{k = 1}^n r_k i_k \right)} + J.$$ Often, the map $\pi$ is omitted, and this ideal is simply written $I(R/J).$ From here, it is straightforward to verify that $I(R/J) = (I + J)/J$ by showing the containments $\subseteq$ and $\supseteq.$
Given commutative unital rings $R$ and $S$ such that there exists a ring homomorphism $\varphi : R \to S,$ we define the extension of an ideal $I$ of $R$ to be the ideal $$I^e = \varphi(I) S = {\left \{\sum_{k = 1}^n s_k \varphi(i_k) \,|\, n \geq 0 \text{ is an integer, } i_k \in I, \text{ and } s_k \in S \text{ for all } k \right \}}$$ of $S.$ For an ideal $J$ of $R,$ there is a ring homomorphism $\pi : R \to R/J$ defined by $\pi(r) = r + J.$ Putting this together with the above shows that $I^e = \pi(I)(R/J)$ consists of the elements $$\sum_{k = 1}^n (r_k + J)(i_k + J) = \sum_{k = 1}^n (r_k i_k + J) = {\left(\sum_{k = 1}^n r_k i_k \right)} + J.$$ Often, the map $\pi$ is omitted, and this ideal is simply written $I(R/J).$ From here, it is straightforward to verify that $I(R/J) = (I + J)/J$ by showing the containments $\subseteq$ and $\supseteq.$