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About the following theorem:

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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 $Im(M) \cong R/Ann()$ holds but now $Im(M) \ne M$ if $m \ne 0$.

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    I may be a little rusty on this topic, but the proposition requires that $\langle m \rangle$ is a cyclic $R$-module. In that case could it be true that $am = 0$ for all $a\in R$? – 2017-01-18
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    A cyclic module $M = \langle m\rangle$ means that $f$ is surjective by definition. – 2017-01-18
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    @DavidK I cant understand why not possible that am=0 for all a∈R even if is a cyclic R-module ? Thank you – 2017-01-18
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    @PrahladVaidyanathan. Sorry I dont understand why. – 2017-01-18
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    @Edi: By definition, $\langle m\rangle = \{am : a\in R\}$ – 2017-01-18
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    @PrahladVaidyanathan. Ha! I thought $\langle m\rangle $ ıs a group cyclic. Thanks – 2017-01-18
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    @DavidK This is possible, but you'd have to be working with rings not having unity. Typically the requirement that $1\cdot m=m$ prevents the annihilator from consuming the whole ring. – 2017-01-18

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You have that $M$ is generated by the element $m$ i.e $M==\{a.m : a\in R\}$. 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)$