These are the propositions in question.
I can't understand the observations for $\alpha(M/N) = (\alpha M+N)/N$ and $N+mM = M$. In general how do we prove observations like this. I can't think beyond looking for a homomorphism and then quotienting with the kernel.
Thanks in advance.
On
To prove $\mathfrak{a}(M/N)=(\mathfrak{a}M+N)/N:$
Let $\sum_{i=1}^n a_i(x_i+N)\in\mathfrak{a}(M/N)$. Then $$\sum_{i=1}^n a_i(x_i+N)=\sum_{i=1}^n(a_ix_i+N)=\sum_{i=1}^n a_ix_i+N\in (\mathfrak{a}M+N)/N$$ since $\sum_{i=1}^n a_ix_i\in\mathfrak{a}M\subseteq\mathfrak{a}M+N$. So $\mathfrak{a}(M/N)\subseteq (\mathfrak{a}M+N)/N$. Conversely if $$(\sum_{i=1}^n a_ix_i+p)+N\in(\mathfrak{a}M+N)/N$$ where $\sum_{i=1}^n a_ix_i\in\mathfrak{a}M$ and $p\in N$ then $$(\sum_{i=1}^n a_ix_i+p)+N=\sum_{i=1}^na_ix_i+(p+N)=\sum_{i=1}^na_ix_i+N=\sum_{i=1}^n(a_ix_i+N)=\sum_{i=1}^na_i(x_i+N)\in\mathfrak{a}(M/N).$$ So $(\mathfrak{a}M+N)/N\subseteq \mathfrak{a}(M/N)$.
To prove $N+\mathfrak{m}M=M:$
Since $N$ and $\mathfrak{m}M$ are both submodules of $M$, so $N+\mathfrak{m}M\subseteq M$. Also if $x\in M$, then $x+\mathfrak{m}M\in M/\mathfrak{m}M$ so that $$x+\mathfrak{m}M=\sum_{i=1}^n (a_i+\mathfrak{m})(x_i+\mathfrak{m}M)=\sum_{i=1}^n(a_ix_i+\mathfrak{m}M)=\sum_{i=1}^n a_ix_i+\mathfrak{m}M$$ (where $a_i+\mathfrak{m}\in A/\mathfrak{m}$, $1\leq i\leq n$) since $x_i+\mathfrak{m}M$ generates $M/\mathfrak{m}M$. Therefore $$x-\sum_{i=1}^na_ix_i\in\mathfrak{m}M\Rightarrow x-\sum_{i=1}^na_ix_i=z$$ for some $z\in\mathfrak{m}M\Rightarrow x=\sum_{i=1}^n a_ix_i+z\in N+\mathfrak{m}M$ because $\sum_{i=1}^na_ix_i\in N$ and $z\in\mathfrak{m}M$. Hence $N+\mathfrak{m}M=M$.
First equality:
$\mathfrak a(M/N)$ is the set of finite sums of congruence classes $am+N$ ($a\in\mathfrak a,\;m\in M$), which are in $\mathfrak a M+N$ by definition.
Second equality:
The image of $N$ in $M/P$ (for any submodule $P\subset M$) is $(N+P)/P$, and if the latter is equal to $M/P$, then $N+P=M$.