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.
Write $K\xrightarrow{f}L\xrightarrow{g}M\to 0.$
Take $N=M,$ then $g^*(\text{id}_M)=g \in \text{im } g^*=\ker f^*,$ so $f^*g=gf=0.$
Take $N=L/\text{im} f.$ Let $\pi:L\to N$ be the projection. Then $f^*\pi=\pi f=0,$ so $\pi \in \text{im } g^*,$ i.e. $g^*h=hg=\pi$ for some $h.$ Then $\ker g\subset \ker \pi = \text{im } f.$
Take $N=M/\text{im } g.$ Let $\pi:M\to N$ be the projection. Then $g^*\pi=\pi g=0,$ so $\pi=0$ and therefore $M=\text{im }g.$
Best Answer
To show the reverse direction for both statements, the idea is we have to choose some special module $N$ in statement 1 and $M$ in statement 2; and map $f$ at some point during the proof.
1/ To show $v$ is surjective, we can consider $N=M''/\text{Im}(v)$ and $f$ the projection map $$M\xrightarrow{v}M''\xrightarrow{f} M''/\text{Im}(v)$$ then $f\circ v = 0$. Since this implies $f=0$, we have that $M''/\text{Im}(v)=0$, i.e, $v$ is surjective. The rest of the proof is from Atiyah & MacDonald.
2/ Similarly, to show $u$ is injective, we can consider $M=\text{Ker}(u)$ and $f$ the inclusion map $$\text{Ker}(u) \xrightarrow{f}N' \xrightarrow{u} N$$ then $u\circ f =0$. Since this implies $f=0$, we have that $\text{Ker}(u)=0$ as desired.
To prove $\text{Im}(u) \subset \text{Ker}(v)$, note that $v_* \circ u_* =0$, i.e, $v \circ u \circ f =0$ for all $M\xrightarrow{f}N'$. Choose $M=N'$ and $f$ the identity map, we have $v\circ u =0$, hence $\text{Im}(u) \subset \text{Ker}(v)$.
Lastly, to show that $\text{Ker}(v)\subset\text{Im}(u)$, choose $M=\text{Ker}(v)$ and $f$ the inclusion map $$\text{Ker}(v) \xrightarrow{f}N \xrightarrow{v} N''$$ Since $v\circ f =0$, we have that $f\in \text{Ker}(v_*)=\text{Im}(u_*)$, hence there is a map $g$ $$ \text{Ker}(v) \xrightarrow{g}N' \xrightarrow{u} N $$ such that $u\circ g =f$. Note that $f$ is the inclusion map so this implies that $\text{Ker}(v)\subset\text{Im}(u)$.