[proofplan]
We define addition and scalar multiplication on cosets using representatives in $M$. The main point is well-definedness: changing representatives changes each expression by an element of $N$, and this is exactly where the submodule hypotheses on $N$ are used. Once the operations are well-defined, every module axiom on $M/N$ follows by applying the corresponding module axiom in $M$ and then passing to cosets.
[/proofplan]
[step:Record the coset criterion used to compare representatives]
For $a,b \in M$, the additive cosets $a+N$ and $b+N$ are equal if and only if $a-b \in N$.
Indeed, if $a+N=b+N$, then $a \in b+N$, so there exists $n \in N$ such that $a=b+n$, hence $a-b=n \in N$. Conversely, if $a-b \in N$, then for every $n \in N$ we have
\begin{align*}
a+n=b+\bigl((a-b)+n\bigr).
\end{align*}
Since $N$ is a submodule, it is closed under addition, so $(a-b)+n \in N$. Thus $a+N \subset b+N$. The same argument with $b-a=-(a-b) \in N$ gives $b+N \subset a+N$, so $a+N=b+N$.
[/step]
[step:Show that addition of cosets is independent of representatives]
Suppose $m_1+N=m+N$ and $m_1'+N=m'+N$ for elements $m,m',m_1,m_1' \in M$. By the coset criterion,
\begin{align*}
m_1-m \in N
\end{align*}
and
\begin{align*}
m_1'-m' \in N.
\end{align*}
Since $N$ is a submodule, it is closed under addition, so
\begin{align*}
(m_1+m_1')-(m+m')=(m_1-m)+(m_1'-m') \in N.
\end{align*}
Applying the coset criterion again gives
\begin{align*}
(m_1+m_1')+N=(m+m')+N.
\end{align*}
Therefore the rule $(m+N)+(m'+N)=(m+m')+N$ is well-defined.
[guided]
We must prove that the formula for adding cosets does not depend on the chosen representatives. Suppose the same two cosets are represented in two different ways:
\begin{align*}
m_1+N=m+N
\end{align*}
and
\begin{align*}
m_1'+N=m'+N.
\end{align*}
By the coset criterion, changing the representative from $m$ to $m_1$ means exactly that the difference $m_1-m$ lies in $N$, and changing the representative from $m'$ to $m_1'$ means exactly that $m_1'-m'$ lies in $N$.
Now compare the two possible sums of representatives. Their difference is
\begin{align*}
(m_1+m_1')-(m+m')=(m_1-m)+(m_1'-m').
\end{align*}
Both terms on the right-hand side belong to $N$, and $N$ is closed under addition because $N$ is a submodule. Hence this whole difference belongs to $N$. Applying the coset criterion once more gives
\begin{align*}
(m_1+m_1')+N=(m+m')+N.
\end{align*}
Thus the sum of cosets is independent of the representatives chosen, which is precisely the well-definedness of addition on $M/N$.
[/guided]
[/step]
[step:Show that scalar multiplication of cosets is independent of representatives]
Suppose $m_1+N=m+N$ for elements $m,m_1 \in M$, and let $r \in R$. By the coset criterion, $m_1-m \in N$. Since $N$ is closed under scalar multiplication by elements of $R$,
\begin{align*}
rm_1-rm=r(m_1-m) \in N.
\end{align*}
Again by the coset criterion,
\begin{align*}
rm_1+N=rm+N.
\end{align*}
Therefore the rule $r(m+N)=rm+N$ is well-defined.
[/step]
[step:Verify the abelian group structure on the quotient]
Let $0_M$ denote the additive identity of $M$, and let $1_R$ denote the multiplicative identity of $R$. Let $x,y,z \in M/N$. Choose $m,m',m'' \in M$ such that $x=m+N$, $y=m'+N$, and $z=m''+N$. Since addition is well-defined, the following computations may be made with these representatives.
Associativity follows from associativity in the additive group of $M$:
\begin{align*}
(x+y)+z=((m+m')+m'')+N=(m+(m'+m''))+N=x+(y+z).
\end{align*}
Commutativity follows from commutativity in the additive group of $M$:
\begin{align*}
x+y=(m+m')+N=(m'+m)+N=y+x.
\end{align*}
The zero element is $0_M+N=N$, because
\begin{align*}
x+(0_M+N)=(m+0_M)+N=m+N=x.
\end{align*}
The additive inverse of $x=m+N$ is $(-m)+N$, because
\begin{align*}
x+((-m)+N)=(m-m)+N=0_M+N.
\end{align*}
Thus $M/N$ is an abelian group under the induced addition.
[/step]
[step:Verify the scalar multiplication axioms]
Let $r,s \in R$ and let $x,y \in M/N$. Choose $m,m' \in M$ such that $x=m+N$ and $y=m'+N$. Using the module axioms in $M$, we obtain
\begin{align*}
r(x+y)=r((m+m')+N)=(r(m+m'))+N=(rm+rm')+N=rx+ry.
\end{align*}
Similarly,
\begin{align*}
(r+s)x=((r+s)m)+N=(rm+sm)+N=rx+sx.
\end{align*}
The compatibility of ring multiplication with scalar multiplication is
\begin{align*}
(rs)x=((rs)m)+N=(r(sm))+N=r(sx).
\end{align*}
Finally, since $M$ is a unital left $R$-module,
\begin{align*}
1_R x=(1_Rm)+N=m+N=x.
\end{align*}
Therefore the induced scalar multiplication satisfies all left $R$-module axioms.
[/step]
[step:Conclude that the quotient is a left module]
The operations on $M/N$ are well-defined by the preceding well-definedness steps. The induced addition makes $M/N$ an abelian group, and the induced scalar multiplication satisfies the distributive, associative, and identity axioms for a left $R$-module. Hence $M/N$ is a left $R$-module under the operations inherited from $M$.
[/step]
