About the following theorem:
I cannot understand Why $f$ is a surjective function?
Naively speaking, since for any $a.m$ there is an $a\in R$ so we are done! But the set $M'={\{a.m \ | \ a\in R \ , \ m\in M }\}$ may or maynot equal to $M$. For example, define $f(a)=0$ for any $a\in R$ and still $Im(M) \cong R/Ann(<m>)$ holds but now $Im(M) \ne M$ if $m \ne 0$.
You have that $M$ is generated by the element $m$ i.e $M=<m>=\{a.m : a\in R\}$. When you define the function $f:R\rightarrow M$ you send each element $a\in R$ to $a.m$ the image of all the elements of $R$ is the set $M$, so f is surjective.
Besides, $ker(f)=\{a\in R : f(a)=0\}=\{a \in R : a.m=0\}=Ann(m).$ As $f$ is homomorphism (it´s evident) you have than by the first isomorphism theorem : $ R/ker(f)=R/Ann(m)\simeq M=Im(f)$