Showing that $\mathrm{Hom}_R(R/I, M) \cong \mathrm{Ann}_I(M)$ 
$R$ is a ring (not necessarily commutative or with identity), $I$ is an ideal of $R$, and $M$ is an $R$-module. Also $\mathrm{Ann}_I(M) = \{m \in M \mid \forall i \in I \ im = 0\}$.
I want to show that $\mathrm{Hom}_R(R/I, M) \cong \mathrm{Ann}_I(M)$.

I saw that this question was asked here, but it didn't really help me.
To make things easier I first tried to work with rings with identity.
Let's say we have $\phi \in \mathrm{Hom}_R(R/I, M)$ then $\phi(r + I) = \phi(r(1 + I)) = r\phi(1 + I)$ so $\phi$ is fully determined by $\phi(1 + I)$. But all this does is give me a map $\mathrm{Hom}_R(R/I, M) \to M$. I don't know how to show that $m \in \mathrm{Ann}_I(M)$. I'm not sure if this approach would work and even if it does, I don't know what I should do for rings without identity.
This is just an idea but I was thinking about starting from $\mathrm{Hom}_R(R, M)$, maybe finding a map $f: \mathrm{Hom}_R(R, M) \to M$ such that $\mathrm{Hom}_R(R, M)/\ker(f) \cong \mathrm{Hom}_R(R/I, M) \cong \mathrm{Im}(f) = \mathrm{Ann}_I(M)$ but I have made no progress on this.
Hints would be appreciated. Thanks in advance.
 A: Here is another approach.
If $M$ is an $R$-module, and $N$ is a submodule of $M$, when we want to define a homomorphism with domain $M/N$, is sufficient define a homomorphism with domain $M$ whose kernel contains $N$. Explicitly, if $P$ is an $R$-module, and $\alpha \colon M \to P$ is a homomorphism with $N \subseteq \ker \alpha$, then we can define $\widetilde\alpha \colon M/N \to P$ via $\widetilde\alpha(m+N) := \alpha(m)$ for all $m \in M$. Sometimes, this is stated (in a very little enlightening way) as follows:

If $P$ is an $R$-module, then for every homomorphism $\alpha \colon M \to P$ with $N \subseteq \ker \alpha$ there exists a unique homomorphism $\widetilde\alpha \colon M/N \to P$ such that $\widetilde\alpha \circ \pi = \alpha$, where $\pi \colon M \to M/N$ is the canonical quotient map.

Note that the words "for every ... there exists a unique" is a fancy way to say that the function
$$
\begin{align*}
\operatorname{Hom}_R(M/N,P) & \longrightarrow \{\alpha \in \operatorname{Hom}_R(M,P): N \subseteq \ker \alpha\} \\
\beta & \longmapsto \beta \circ \pi
\end{align*}
$$
is bijective.$^1$ Futhermore, one can check that the set in the right is a subgroup of $\operatorname{Hom}_R(M,P)$, and that above function is a group homomorphism, hence a group isomorphism.
One more thing that we need to remember is the group isomorphism $\operatorname{Hom}_R(R,M) \cong M$ that identifies $\alpha \in \operatorname{Hom}_R(R,M)$ with $\alpha(1) \in M$.
As an exercise, I will left to you to think about the following chain of group isomorphisms:
$$
\begin{align*}
\operatorname{Hom}_R(R/I,M) & \cong \{\alpha \in \operatorname{Hom}_R(R,M) : I \subseteq \ker \alpha\} \\
&= \{\alpha \in \operatorname{Hom}_R(R,M): \forall i \in I,\, \alpha(i)=0\} \\
&= \{\alpha \in \operatorname{Hom}_R(R,M): \forall i \in I,\, i\alpha(1)=0\} \\
&\cong \{m \in M : \forall i \in I,\, im=0\} \\
&= \operatorname{Ann}_I(M).
\end{align*}
$$

$^1$ A function $f \colon X \to Y$ is bijective if for every $y \in Y$ there exists a unique $x \in X$ such that $f(x)=y$, isn't?
A: Note: this is only for unital rings as it explicitly uses $1 \in R$.
Define the map
$$
\DeclareMathOperator{\Hom}{Hom} \DeclareMathOperator{\Ann}{Ann} 
\begin{array}{rccc} 
\alpha: & \Hom_R(R/I, M) & \longrightarrow & \Ann_I(M) \\
& \phi & \longmapsto & \phi(1+I)
\end{array}
$$
then show that $\alpha$ is

*

*a well-defined homomorphism of abelian groups:
$\quad \alpha(\phi_1 + \phi_2) = \alpha(\phi_1) + \alpha(\phi_2)$

*surjective: $\quad\operatorname{Im} \alpha = \Ann_I(M)$

*injective: $\quad\operatorname{Ker} \alpha = \{0\}$
Along with the hints in the comments, can you take it from here?
