Let $R$ and be a (unital) commutative ring with $A$ as an $R$-module. Now suppose $f\in{Hom(R,f(R))}$ I am wondering what requirements (if any exist) need to be placed on $f$ to ensure that $A$ is an $f(R)$- module. My intuition told me that $A$ is an$f(R)$module if $R\cong{f(R)}$ since we can consider $R$ and $f(R)$ equivalent. After considering that the only difference between $f$ as a ring homomorphism and as an isomorphism is the requirements of injectivity and surjectivity, I concluded that only surjectivity be required of the homomorphism as follows: if we assume $f$ is surjective then $\forall{s\in{f(R)}}$ we can write $s$ as $f(r)$ for some $r\in{R}$ and thus write the axioms for $A$ being an $f(R)$- module as $(f(r_1)+f(r_2))a=f(r_1)a+f(r_2)a$
$f(r_1)(a_1+a_2)=f(r_1)a_1+f(r_1)a_2$
$(f(r_1)f(r_2))a=f(r_1)(f(r_2)a)$
$f(1_R)a=a$
Since surjectivity is needed to write the above axioms of $A$ being an $f(R)$-module in terms of the elements of $f(R)$ as images elements of $R$ under $f$, I see no reason to require injectivity of $f$ so I am asserting that $f$ only needs to be a surjective homomorphism to ensure $A$ is an $f(R)$-module. Any confirmation or refutation would be appreciated.
A function $f$ is always surjective onto its image!
The question is a little imprecise. I presume, for instance, that you want $A$ to be an $f(R)$-module in a unique and natural way (i.e. coming as a consequence of the $R$-module structure). Consider the following example:
$$R = \mathbb{Z}\oplus\mathbb{Z},$$ $$A = \mathbb{Z}x\oplus\mathbb{Z}y = \{\lambda x + \mu y | \lambda, \mu \in \mathbb{Z}\}$$
where $A$ is an $R$-module under the action $(\lambda,\mu)\cdot x = \lambda x$ and $(\lambda,\mu)\cdot y = \mu y$.
Now let $f$ be the function $f: R \to \mathbb{Z}, f(\lambda,\mu) = \lambda$. But $A$ isn't an $f(R)$-module in an obvious way: for instance, $(\lambda,0)\cdot y \neq (\lambda,1)\cdot y$, so how do we define $f(\lambda,0)\cdot y$, given that the answer has to be the same as $f(\lambda,1)\cdot y$, because both are $\lambda \cdot y$?
(Let's not confuse this with something similar: $f(R)$ is isomorphic to $\mathbb{Z}$, and $A$ is certainly a $\mathbb{Z}$-module in many ways. But saying that you want it to be "an $f(R)$-module" usually implies that you want the module structure to be compatible with $f$ and $R$ somehow - as detailed below.)
To make $A$ an $f(R)$-module in a unique and natural way, you need not only that $A$ should be an $f(R)$-module, but also that $f(R)$ should act on $A$ in the same way that $R$ acts on $A$. That is, given $r\in R$ and $a\in A$, we should have $r\cdot a = f(r)\cdot a$. (If you like, you can view this as the definition of $f(r)\cdot a$ (the new $f(R)$-module structure on $A$) in terms of $r\cdot a$ (the old $R$-module structure on $A$).
The only obstruction to this, really, is if $f(r) = f(s)$ but $r\cdot a \neq s\cdot a$ for some $a\in A$ and $r\neq s \in R$.
You should be able to convince yourself that a necessary and sufficient condition for this desired module structure to exist is: for all $r\in \ker f$ and all $a\in A$, $r\cdot a = 0$.