Showing there is an exact sequence Consider the following commutative diagram with exact rows (of $R$-modules and $R$-linear maps):
$$
\newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex}
\newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}}
\begin{array}{c}
0 & \ra{} & M' & \ra{f} & M & \ra{g} & M'' & \ra{} & 0\\
  &       &\da{\alpha'} &&\da{\alpha} &&\da{\alpha''}\\
0 & \ra{} & N' & \ra{\smash{f'}} & N & \ra{\smash{g'}} & N'' & \ra{} & 0
\end{array}
$$
Suppose $\alpha^{'}$ is an isomorphism. I want to show that there is an exact sequence:
$$0 \longrightarrow M \xrightarrow{\ (g,\alpha)\ } M^{''} \oplus N \xrightarrow{(\alpha^{''},-g')} N^{''} \longrightarrow 0$$
Two questions: to show the last map is surjective, can we simply let $n$ be in $N^{''}$ then since $g^{'}$ is surjective we can find $x \in N$ such that $g(x)=n$. So take $(0,-x) \in M^{''} \oplus N$. 
Question 2: I need to show that the image of the first (nonzero map) is equal to the kernel of $(\alpha^{''},-g')$. I'm stuck in showing that the kernel is contained in the image, can you please help with this part?
 A: *

*Yes; what you state would show that the map $(\alpha'',-g')$ is onto, except for the typo: you used $g$ when you meant $g'$.

*Suppose $(m'',n)$ lies in the kernel. That means that $\alpha''(m'')=g'(n)$. Since $g$ is onto, there exists $m\in M$ such that $g(m)=m''$, so 
$$g'(n) = \alpha''(m'') = \alpha''(g(m)) = g'(\alpha(m)).$$
Therefore, $n-\alpha(m)\in\mathrm{ker}(g')$; therefore, there exists $n'\in N'$ such that $f'(n') = n-\alpha(m)$. Let $m'\in M'$ correspond to $n'$. Then 
$$\alpha(f(m')) = f'(\alpha'(m')) = f'(n') = n-\alpha(m).$$
Therefore, $\alpha(f(m')+m) = n$. So 
$$(g,\alpha)(m,f(m')+m) = (g(m),\alpha(f(m')+m)) = (m'',n),$$
so $(m'',n)\in\mathrm{Im}(g,\alpha)$.
Added. Note that we only require $\alpha'$ to be onto (not necessarily an isomorphism); this guarantees the existence of $m'\in M'$ with $\alpha'(m') = n'$, which suffices for the inclusion here.

*Now, if $m\in M$, then
$$(\alpha'',-g')\bigl((g,\alpha)(m)\bigr) = (\alpha'',-g')(g(m),\alpha(m)) = \alpha''(g(m)) - g'(\alpha(m)) = 0$$
so $\mathrm{Im}(g,\alpha)\subseteq \mathrm{ker}(\alpha'',-g')$. 
