Well, your proof is okay. Let me suggest a slightly different way of looking at it:
Consider a sequence
$$ 0\; \xrightarrow{\phantom{ij}} \; A \; \xrightarrow{i\phantom{j}}\; B\; \xrightarrow{j\phantom{i}} \; C$$
and look at
$$ 0\; \xrightarrow{\phantom{j_{\ast}}}
\operatorname{Hom}{(M,A)}\; \xrightarrow{i_{\ast}}\operatorname{Hom}{(M,B)} \;\xrightarrow{j_{\ast}}\;
\operatorname{Hom}{(M,C)},$$
where I write $i_{\ast} = \operatorname{Hom}(M,i)$ and $j_{\ast} = \operatorname{Hom}(M,j)$.
Saying that the first sequence is exact amounts to saying $i = \ker{j}$, that is $ji = 0$ and $i$ has the universal property as depicted in the first diagram below: If $g: M \to B$ is such that $jg = 0$ then there exists a unique $f: M \to A$ such that $if = g$. In other words if $j_\ast g = 0$ then $g = i_{\ast}f$, or yet again $\operatorname{Ker}{j_\ast} \subset \operatorname{Im}{i_{\ast}}$ and $i_{\ast}$ is injective.
On the other hand, the second diagram says: if $g: M \to B$ is of the form $g = if = i_{\ast}f$ then $j_{\ast}g = 0$ (because $j_\ast g = j_{\ast}i_{\ast} f = (ji)_{\ast}f = 0f = 0$). In other words, $\operatorname{Im}{i_{\ast}} \subset \operatorname{Ker}{j_{\ast}}$.
Summing up, we have shown that for all $M$ the sequence
$$ 0\; \xrightarrow{\phantom{j_{\ast}}}
\operatorname{Hom}{(M,A)}\; \xrightarrow{i_{\ast}}\operatorname{Hom}{(M,B)} \;\xrightarrow{j_{\ast}}\;
\operatorname{Hom}{(M,C)}$$
is exact both at $\operatorname{Hom}{(M,A)}$ ($i_{\ast}$ is injective) and at $\operatorname{Hom}{(M,B)}$ ($\operatorname{Im}{i_{\ast}} = \operatorname{Ker}{j_{\ast}}$)—you seem to have forgotten about the first point here.
Added: As witnessed by the argument above, left exactness of $\operatorname{Hom}$ is essentially the definition of left exactness in the abelian category of $R$-modules. As the comments try to point out, the importance of this fact cannot be overemphasized.
I would like to add two further points:
A functor $F$ is left exact in your definition if and only if $0 \to F(A) \to F(B) \to F(C)$ is left exact for every short exact sequence $0 \to A \to B \to C \to 0$.
Indeed, in a left exact sequence $0 \to A \to B \to C$, we may factor $j: B \to C$ over its image as $B \twoheadrightarrow \operatorname{Im}{j} \rightarrowtail C$ and obtain two exact sequences $$0 \to A \to B \to \operatorname{Im}{j} \to 0 \qquad \text{and} \qquad 0 \to \operatorname{Im}{j} \to C \to \operatorname{Coker}{j} \to 0.$$ Applying $F$ to these two exact sequences, we obtain the left exact sequences $$0 \to F(A) \to F(B) \to F(\operatorname{Im}{j}) \qquad \text{and} \qquad 0 \to F(\operatorname{Im}{j}) \to F(C ) \to F(\operatorname{Coker}{j}).$$ Since the kernel of a map is not changed by postcomposing the map with a monomorphism (check this!), we have $$\operatorname{Ker}{(F(B) \to F(\operatorname{Im}{j}))} = \operatorname{Ker}{(F(B) \to F(\operatorname{Im}{j}) \to F(C))},$$ so by functoriality of $F$ we get a left exact sequence $0 \to F(A) \to F(B) \to F(C)$ as desired.
A natural question is: When does $\operatorname{Hom}(M,-)$ send short exact sequences to short exact sequences? In other words, when is $j_\ast = \operatorname{Hom}{(M,j)}$ an epimorphism for all short exact sequences $0\; \xrightarrow{\phantom{ij}} \; A \; \xrightarrow{i\phantom{j}}\; B\; \xrightarrow{j\phantom{i}} \; C \to 0$?
In view of left exactness of $\operatorname{Hom}{(M,-)}$ the question is: Given any morphism $h: M \to C$ and any epimorphism $j: B \to C$, when is $h$ of the form $h = j_\ast g$ for some morphism $g: M \to B$?
As you certainly know, this is precisely the definition of projective modules: $M$ is called projective if and only if $g$ always exists, for all epimorphisms $j: B \twoheadrightarrow C$ and all $h: M \to C$. For emphasis:
A module $M$ is projective if and only if $\operatorname{Hom}{(M,-)}$ is exact, that is: it sends short exact sequences to short exact sequences.
Generally looks good, though as Berci says in the comments you should check that $\phi$ doesn't depend on your choices.
However, I write an answer, because I would suggest an alternative method of proof for
$\newcommand\im{\operatorname{im}}\ker\alpha^*=\im\beta^*$.
The key is to notice that $C\cong B/\im\alpha$, and $\beta : B\to C$ is the quotient map.
Therefore you can use the universal property of the quotient, which is that maps $\psi : B\to M$ such that $\psi(\im\alpha)=0$ are in one-to-one correspondence with maps $\tilde{\psi} : C\to M$, and the correspondence is given by $\psi = \tilde{\psi}\circ \beta$.
Then $\psi(\im\alpha)=0$ if and only if $\alpha^*\psi = \psi\circ \alpha =0$ if and only if $\psi\in \ker\alpha^*$. Thus $\psi\in\ker\alpha^*$ if and only if
$\psi=\tilde{\psi}\circ \beta$ for some $\tilde{\psi}\in \operatorname{Hom}_R(C,M)$, i.e., $\psi \in \ker\alpha^*$ if and only if $\psi \in \im\beta^*$, as desired.
Best Answer
Say that two elements $f,g\in {\rm Hom}_R(D,M)$ are equivalent ($f\sim g$) if the image of $f-g$ lies in the image of $\psi$. Let $K$ denote a complete set of representatives of equivalence classes. Then we may bijectively identify: $${\rm Hom}_R(D,M)\cong K \times {\rm Hom}_R(D,{\rm im}(\psi))$$
Exactness of the Hom sequence is by definition equivalent to the following two maps being bijective:
(1) The map $\varphi'\colon K \to {\rm Hom}_R(D,N)$
(2) The map $\psi'\colon{\rm Hom}_R(D,L) \to {\rm Hom}_R(D,{\rm im{(\psi)}}) $.
Thus we may define a relation between ${\rm Hom}_R(D,M)$ and ${\rm Hom}_R(D,N)\times {\rm Hom}_R(D,L)$ which is bijective if and only if the Hom sequence is exact.