$\text{Hom}$ functor reflects left/right exactness 

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*Let $M'\xrightarrow{u}M\xrightarrow{v}M''\to 0$ be a sequence of $A$ module homomorphisms. This sequence is exact if and only if for all $A$-module $N$, $0\to\text{Hom}(M'',N)\xrightarrow{v^*}\text{Hom}(M,N)\xrightarrow{u^*}\text{Hom}(M',N)$ is exact.

*Let $0\to N'\xrightarrow{u}N\xrightarrow{v} N''$ be a sequence of $A$-modules and homomorphisms. This sequence is exact if and only if for all $A$-modules $M$, the sequence $0\to\text{Hom}(M,N')\xrightarrow{u_*}\text{Hom}(M,N)\xrightarrow{v_*}\text{Hom}(M,N'')$ is exact.



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*For $(\Leftarrow)$, A&M actually mentioned $v^*$ is injective then $v$ is surjective as an obvious fact. Well, if $f\in\text{Hom}(M'',N)$ such that $f\circ v = 0$ implies $f =0$ so maybe the range of $v$ should be the whole $M''$. But I don't know the exact proof of this.

*For $(\Leftarrow)$, same as before, intuitively, if $f\in\text{Hom}(M,N')$ such that $u_*(f) = u\circ f=0$ then $f=0$, maybe $u$ should be injective. But similarly I don't know the exact proof of this. Also, how can I prove $\text{Ker}(v)\subset\text{Im}(u)$? In 1, A&M set $N = M/\text{Im}(u)$ but in this case, I don't know how to set $M$. Could you help?

Note. Most of the proof given in website shows $(\Rightarrow)$ direction which is fairly easy. My question is the reverse direction.
 A: 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 $\text{Ker}(v)\subset\text{Im}(u)$.
A: Q1. You want to check that $v$ is onto and that $\ker v = \operatorname{im} u $. To do this, it suffices to show first that if $f: M''\to N$ is any $A$-linear map with $f\circ v=0$ then $v=0$, and this is exactly what injectivity of $v^*$ is stating.
On the other hand, you know that $u^*v^*(f) = f\circ v\circ u=0$ for any $f$, so taking $f=1_{M''}$ gives $v\circ u=0$ i.e $\operatorname{im} u \subseteq \ker v$.
To check the other inclusion, note that the map $\pi: M\to M/\operatorname{im} u $ is such that $u^*(\pi)=0$ so there must be $\psi$ such that $\pi = v^*(\psi) = \psi\circ v$. This means that if $vx=0$ then $x\in\ker\pi = \operatorname{im} u$.
Q2. For the converse, choose $A=M$, and let $\mathcal S$ be your short exact sequence. Then $\hom_A(A,\mathcal S)$ is naturally isomorphic to $\mathcal S$, since $\hom_A(A,-)$ is naturally isomorphic to the identity functor, so if $\hom_A(A,\mathcal S)$ is exact so is $\mathcal S$.
