The following is taken from "Modules an approach to linear algebra" by Blyth
$\color{Green}{Background:}$
$\textbf{Exercise:}$ Let $M$ be an $R-$module. If $r,s\in R$ show that
$$r-s\in \mathrm{Ann}_{R}S\Longrightarrow (\forall x\in M) rs=sx.$$
Deduce that $M$ can be considered as an $R/\mathrm{Ann}_{R}M-$module. Show that the annihilator of $M$ in $R/\mathrm{Ann}_{R}M$ is zero.
$\textbf{[Solution:]}$
Let $r-s\in \mathrm{Ann}_{R}M,$ then $(r-s)x=0$ for all $x\in M,$ which implies $rx=sx.$
To show that $M$ can be considered as a $R/\mathrm{Ann}_{R}S$ module. Any element of $R/\mathrm{Ann}_{R}M$ can be written as $r' + 1_R m=r'+ m \in r'+ \mathrm{Ann}_{R}M$ for some $m\in M,$ and $1_R\in R.$ And $1_R\in R$ since $M$ is considered a $R-$module by hypothesis. Then for any $\lambda, \mu\in R/\mathrm{Ann}_{R}M, x.y\in M,$ we have
$\lambda x + R/\mathrm{Ann}_{R}M$ and $\lambda y + R/\mathrm{Ann}_{R}M,$ and $(\lambda x + R/\mathrm{Ann}_{R}M)+(\lambda y + R/\mathrm{Ann}_{R}M)=(\lambda x+\lambda y)+R/\mathrm{Ann}_{R}M=(\lambda(x+y))+R/\mathrm{Ann}_{R}M,$
$[(\lambda + R/\mathrm{Ann}_{R}M)+(\mu+R/\mathrm{Ann}_{R}M)]m=((\lambda+\mu) + R/\mathrm{Ann}_{R}M)m=((\lambda+\mu)m) + R/\mathrm{Ann}_{R}M=(\lambda m+\mu m) + R/\mathrm{Ann}_{R}M,$
$(\lambda + R/\mathrm{Ann}_{R}M)(\mu + R/\mathrm{Ann}_{R}M)=\lambda\mu + \mathrm{Ann}_{R}M,$ and for any $x\in M,$ $(\lambda\mu + \mathrm{Ann}_{R}M)x=(\lambda\mu)x + \mathrm{Ann}_{R}M=\lambda(\mu x) + \mathrm{Ann}_{R}M.$
and for all $x\in M, 1_R+R/\mathrm{Ann}_{R}M x= 1_R x + R/\mathrm{Ann}_{R}M=x+R/\mathrm{Ann}_{R}M.$ Hence $M$ is a $R/\mathrm{Ann}_{R}M-$module.
$\color{Red}{Questions:}$
I would like if someone can tell me if I written out the solution with the correct notation, especily about showing that $M$ is a $R/\mathrm{Ann}_{R}M-$module. Specifically, I am not sure if I am allow the operation such as $\lambda + R/\mathrm{Ann}_{R}M)m=\lambda m + R/\mathrm{Ann}_{R}M.$
As for the last part of the question where I have to show that the anniihilator of $M$ in $R/\mathrm{Ann}_{R}M$ is zero.
A friend told me the following:
"$q\neq 0\in R/\mathrm{Ann}_{R}M \Longleftrightarrow q\neq \mathrm{Ann}_{R}M \Longleftrightarrow \exists m\in M, qm\neq 0.$"
I have a question about what my friend told me, specifically about his notations. $q\neq 0\in R/\mathrm{Ann}_{R}M$ would mean that $q\neq 0+rx\in \mathrm{Ann}_{R}M,$ for some $r\in R,$ and all $x\in M.$ Then how does that imply $q\neq \mathrm{Ann}_{R}M \Longleftrightarrow \exists m\in M, qm\neq 0?$
The one big problem with the solution you've given is that you are verifying the properties of a module action without showing the action you're using is well-defined. That is what the condition about the annihilator is all about.
It is specifically explained here.
Another big thing hindering you is a lot of confusion as to what some things mean and their notations. For example
There is no need, and indeed it is sometimes fatally misleading, to talk about elements of $I$ when considering a coset $r+I$. The thing $r+I$ is an element, and there isn't anything internal you need from $I$ to write it. For another thing with this example, writing $m$ would be misleading in any case since $\mathrm{Ann}_R(M)\subseteq R$. It looks like you think it will be in $M$.
Another example:
$\lambda x$ is an element of $M$ and $R/\mathrm{Ann}_RM$ is a quotient ring and the "$+$" between them makes no sense. I think you are trying to write the following:
$$(\lambda +\mathrm{Ann}_RM)x + (\lambda +\mathrm{Ann}_RM)y \\ =\lambda x+\lambda y=\lambda(x+y)\\ =(\lambda +\mathrm{Ann}_RM)(x+y)$$
which is indeed the case (if you've shown the module action is well-defined.)
If you understand quotients and what the annihilator is, then the last part solves itself, and there's nothing mysterious about it. To fix the clarity issues in the following:
I would write
$$q+\mathrm{Ann}_{R}M\neq 0+\mathrm{Ann}_{R}M \\ \Longleftrightarrow q\notin \mathrm{Ann}_{R}M \\ \Longleftrightarrow \exists m\in M, qm\neq 0.$$
What's important here is that this last figure $qm=(q+\mathrm{Ann}_RM)m$ in the proposed module action. This would justify that the only thing which annihilates every $m\in M$ is just $0+\mathrm{Ann}_RM$, which means $M$ is a faithful $R/\mathrm{Ann}_RM$ module.
My advice would be to slow down, take your time to properly understand the notation you are writing, and frankly maybe stop using so much notation and swap it for words. It's not uncommon early on to get caught up in long chains of symbolic reasoning, and as we can see here overemphasis on that can lead to confusion and misunderstanding. Using words usually helps to dispel that problem.