Let $M$ be a Abelian group and $G$ be cyclic group of order $2$. Let $M$ be a $G$-module. Suppose $G=\langle\sigma\rangle.$ Let $M^{G} =\{m\in M\mid \sigma m =m\}$. Define the norm map $N: M\rightarrow M^{G}$ by $N(m)=m+\sigma m$. We want to prove $H^{2}(G,M) \cong M^{G}/im(N)$.
If $m\in M^{G}$ Let $f_{m}$ be the cochain given by $$f_{m}(\sigma^{i}, \sigma^{j})=\left\{\begin{array}{cc} 1, & \quad \text{ if $i+j < 2$}\\ m, & \quad \text{ if $i+j \geq 2$} \end{array} \right. $$ $f_{m}$ is known to be a cocycle.
Thus we can define a map $\phi: M^G\to Z^2(G,M)$.
$\phi$ is injective. This is because $f_a(1,\sigma)=f_a(\sigma,1)=f_{\sigma}(1,1)=0$.
But why is this map surjective ?
What an element of $M^G$ which sends to an arbitrary element of $Z^1(G,M)$?
Before I start, I should say this is not the best way to do things. The computation of $H^n(G;M)$ for $G$ cyclic is quite easy using more abstract machinery, see chapter $6$ of Weibel. But if we want to get our hands dirty... it is possible in this simple case.
Say $m\in M^G$. You claim (after I fix a few typos): $$\phi_m:\begin{pmatrix}1,1\\\sigma,1\\1,\sigma\\\sigma,\sigma\end{pmatrix}\to\begin{pmatrix}0\\0\\0\\m\end{pmatrix}$$Defines a $2$-cocycle. The only nontrivial check would be for $f=g=h=\sigma$ and in this case $\sigma\cdot m-0+0-m=0$ as required, noting $\sigma^2=1$. This crucially relies on $m\in M^G$.
But is the cohomology class $[\phi_m]\in H^2(G;M)$ well defined for $[m]\in M^G/N(M)$? Clearly $m\mapsto\phi_m$ is a group homomorphism. Therefore we need only check that if $m=\sigma n+n$ for some $n$, then $\phi_n$ is a coboundary. Indeed, $\beta(1):=0,\beta(\sigma):=n$ works (check it). Only now do I agree we have a well defined homomorphism $M^G/N(M)\to H^2(G;M)$ (in your previous post about $H^1$ you also forgot to do the necessary well-definedness checks - be careful!).
We want to check injectivity of this map. Note it is NOT true that injectivity of $M^G\to Z^2$ implies (or is implied by) injectivity of $M^G/N(M)\to Z^2/B^2$, in general, so I'd say you're focussing on the wrong thing.
Say $m$ has $\phi_m$ a coboundary of some $\beta$. By setting $g=1$ we conclude $\beta(1)=0$. By setting $\sigma=g=h$ we conclude $m=\sigma\beta(\sigma)+\beta(\sigma)=(\sigma+1)\beta(\sigma)=N(n)$ for $n:=\beta(\sigma)$, so $m$ lies in $N(M)$ and $[m]\in M^G/N(M)$ is zero. Hence our map injects.
Now suppose $\phi$ is any $2$-cocycle. Note surjectivity on cohomology is not the same thing as surjectivity on cycles. We want to mess around with the $2$-cocyle functional equation until we find something which is invariant.
By applying $f=g=1$ we see $\phi(1,-)$ is constant; let's call this constant $m_0$. Then we see $f\cdot m_0=\phi(f,1)$ for all $f$ by setting $g=1$. Setting $f=g=h=\sigma$ shows $(\sigma-1)(m_1+m_0)=0$. Let $m:=m_1+m_0$; we see $m\in M^G$. I claim $\phi_m$ is cohomologous with $\phi$. To verify this, we need to produce a $\beta$.
$$\phi(1,1)-\phi_m(1,1)=m_0=1\cdot\beta(1)-\beta(1\cdot1)+\beta(1)=\beta(1)$$Is forced. Then: $$\phi(1,\sigma)-\phi_m(1,\sigma)=m_0=\beta(1)$$Also holds, great. $$\phi(\sigma,1)-\phi_m(\sigma,1)=\sigma\cdot m_0=\sigma\cdot\beta(1)-\beta(\sigma\cdot1)+\beta(\sigma)$$As desired (for any choice of $\beta(\sigma))$.
Final check: $$\phi(\sigma,\sigma)-\phi_m(\sigma,\sigma)=m_1-(m_1+m_0)=-m_0$$And: $$\sigma\beta(\sigma)-\beta(\sigma^2)+\beta(\sigma)=(\sigma+1)\beta(\sigma)-m_0$$Therefore, the only thing we need to do is choose $\beta(\sigma)$ such that $(\sigma+1)\beta(\sigma)=0$. That's easy; we can even take $\beta(\sigma):=0$.
Therefore $\phi-\phi_m$ is a coboundary and $[\phi_m]=[\phi]\in H^2(G;M)$; the map $M^G/N(M)\to H^2(G;M)$ is a well defined injective and surjective homomorphism, hence an isomorphism.