Step in proof that a quotient is isomorphic to cohomology group The following proof is about the cohomology group. But the fact I do not understand is something like, we have two groups $U \cong V$ and two normal subgroups $N \cong M$ in $U$ respectively $V$, what could be said about the quotients. In general this does not imply $U / N \cong V / M$, as for example $G := \mathbb Z_2 \times \mathbb Z_4$ is a counter-example, it has two normal subgroups isomorphic to $\mathbb Z_2$, but one has $\mathbb Z_4$ as factor group, the other $\mathbb Z_2 \times \mathbb Z_2$. What is valid is, that if the isomorphism is implemented by one mapping, that we have isomorphic quotients, i.e. if $\varphi : G \to H$ is injective, then $G / N \cong \varphi(G) / \varphi(N)$.
In the following proof at the end we have two isomorphisms, hence it is concluded that the quotients are isomorphic. As I do not see why this works here I give all the details of the proof. Unfortunately the definitions are a little bit technical.
Let $G$ be a group. A $G$-module is an abelian group $M$ such that
(1) $m\cdot 1 = m$
(2) $m(gh) = (mg)h$
(3) $(m+n)g = mg + ng$
i.e. $G$ acts on $M$ (condition (1) and (2)) and $M$ is an $G$-operator group (condition (3)), i.e. each element gives rise to an automorphism of $M$. 
Define the abelian groups $C^n(G, M)$ by
(1) $C^0(G, M) = M$
(2) $C^n(G, M)$ for $n \ge 1$ is the set of all mappings from the $n$-times cartesian product $G \times \ldots \times G$ to $M$ with pointwise addition.
Then we define mappings $\delta_n \in \mbox{Hom}_{\mathbb Z}(C^n(G, M), C^{n+1}(G, M))$
by
(1) $(f\delta_0)(g) := f - fg$ for $f \in C^0(G, M) = M$,
(2) 
\begin{align*}
 (f\delta_n)(g_1, \ldots, g_{n+1}) & = f(g_2, \ldots, g_{n+1}) \quad + \\
 & + \sum_{i=1}^n (-1)^i f(g_1, \ldots, g_{i-1}, g_i\cdot g_{i+1}, g_{i+2}, \ldots, g_{n+1}) \quad + \\
 & + (-1)^{n+1} f(g_1,\ldots, g_n) g_{n+1}
\end{align*}
for $f \in C^n(G, m)$ with $n \ge 1$. For $n \ge 1$ we set
$$
 H^n(G, M) = \mbox{ker}(\delta_n) / \mbox{im}(\delta_{n-1}).
$$
These are called the cohomology groups. That they are well defined follows by the fact that the above constructed groups with the mappings (i.e. their so called differentials) form a cochain.
Now the proof concerns an alternative description of these cohomology groups. 
Let $\tilde{C}^n(G, m)$ for $n \ge 0$ be the set of mappings from the $(n+1)$-times cartesian product $G \times \ldots \times G$ to $M$ with pointwise addition (i.e. it is an abelian group) and subject to
$$
 f(g_0 g, \ldots, g_n g) = f(g_0, \ldots, g_n)g
$$
for all $g_0, \ldots, g_n, g \in G$. Also we define the mappings
$\tilde{\delta_n} \in \mbox{Hom}_{\mathbb Z}(\tilde C^n(G, M), \tilde C^{n+1}(G, M))$ by
$$
 (f\tilde{\delta_n})(g_0, \ldots, g_{n+1}) = \sum_{i=0}^{n+1} (-1)^i f(g_0, \ldots, \hat{g_i}, \ldots, g_{n+1});
$$
where $\hat{\quad}$ means that the argument is excluded. Then we have
$$
 H^n(G, M) \cong \mbox{ker}(\tilde{\delta_n}) / \mbox{im}(\tilde\delta_{n-1}).
$$

Proof: We define mappings
  $$
 \mu_n \in \mbox{Hom}_{\mathbb Z}(C^n(G, M), \tilde{C^n}(G, M)) \quad\mbox{and}\quad
 \tau_n \in \mbox{Hom}_{\mathbb Z}(\tilde{C^n}(G, M), C^n(G,M))
$$
  by
  \begin{align*}
 (f\mu_0)(g_0) & = f g_0 \quad \mbox{for} \quad f \in C^0(G, M) = M; \\
 g\tau_0 & = g(1) \quad \mbox{for} \quad g \in \tilde{C^0}(G, M);
\end{align*}
  for $n \ge 1$ and $f \in C^n(G, M)$ let
  $$
 (f\mu_n)(g_0, \ldots, g_n) = f(g_0 g_1^{-1}, g_1 g_2^{-1}, \ldots, g_{n-1}g_n^{-1}) g_n;
$$
  for $n \ge 1$ and $h \in \tilde{C^n}(G, M)$ set
  $$
 (h\tau_n)(g_1, \ldots, g_n) = h(g_1 \cdots g_n, g_2 \cdots g_n, \ldots, g_{n-1}g_n, g_n, 1). 
$$
  A trivial computation shows $\mu_n\tau_n = 1$ and $\tau_n \mu_n = 1$. Hence $\mu_n$ is an Isomorphism between the abelian groups $C^n(G,M)$ and $\tilde{C^n}(G, M)$. For $f \in C^n(G, M)$ we have
  \begin{align*}
 (f\mu_n \tilde \delta_n)(g_0, \ldots, g_{n+1})
  & = \sum_{i=0}^{n+1} (-1)^i (f\mu_n)(g_0, \ldots, \hat{g_i}, \ldots, g_{n+1}) \\
 & = \sum_{i=0}^n (-1)^i f(g_0g_1^{-1},\ldots, g_{i-2}g_{i-1}^{-1}, g_{i-1}g_{i+1}^{-1}, g_{i+1}g_{i+2}^{-1}, \ldots, g_n g_{n+1}^{-1}) g_{n+1} \quad + \\
 & + (-1)^{n+1} f(g_0 g_1^{-1}, \ldots, g_{n-1} g_n^{-1}) g_n
\end{align*}
  and 
  \begin{align*}
 (f\delta_n \mu_{n+1})(g_0, \ldots, g_{n+1})
 & = (f\delta_n)f(g_0g_1^{-1}, g_1 g_2^{-1}, \ldots, g_n g_{n+1}^{-1}) g_{n+1} \\
 & = f(g_1 g_2^{-1},\ldots, g_n g_{n+1}^{-1}) g_{n+1} \quad + \\
 & + \sum_{i=1}^n (-1)^i f(g_0 g_1^{-1}, \ldots, g_{i-1}g_{i+1}^{-1}, \ldots, g_n g_{n+1}^{-1}) g_{n+1} \quad + \\
 & + (-1)^{n+1} f(g_0 g_1^{-1},\ldots, g_{n-1}g_n^{-1}) g_n.
\end{align*}
  Hence $\mu_n \tilde{\delta_n} = \delta_n \mu_{n+1}$ and so $\mbox{ker}(\tilde{\delta_n}) = (\mbox{ker}\delta_n)\mu_n$ and $\mbox{im}(\tilde{\delta_n}) = (\mbox{im} \delta_n)\mu_{n+1}$, hence
  $$
  H^n(G, M) = \mbox{ker}(\delta_n) / \mbox{im}(\delta_{n-1})
 \cong \mbox{ker}(\tilde{\delta_n}) / \mbox{im}(\tilde{\delta}_{n-1}).
$$

What I do not understand is the isomorphism stated in the last paragraph. As the isomophisms between the kernel and image are realised by different mappings, I do not see that we can conclude that the quotients are isomorphic. See my introductory remarks?
 A: The key difference between the example that you gave at the beginning and what is going on in the proof is that there are no isomorphisms $\alpha$ and $\beta$ such that the following diagram commutes:
$$
\begin{array}{ccccc}
\mathbb{Z}_2&\rightarrow^d&\mathbb{Z}_2\times\mathbb{Z}_4&\rightarrow&0\\
\downarrow_\alpha&&\downarrow_\beta&&\\
\mathbb{Z}_2&\rightarrow^{d'}&\mathbb{Z}_2\times\mathbb{Z}_4&\rightarrow&0
\end{array}
$$
Above, $d(1)=(1,0)$ and $d'(1)=(0,2)$.
On the other hand, we have the following:
Proposition: Suppose we have subgroups $H\leq G$ and $H'\leq G'$ and there are homomorphisms $\alpha$ and $\beta$ so that the following diagrams commute:
$$
\begin{array}{ccccc}
H&\hookrightarrow&G&\twoheadrightarrow &G/H\\
\downarrow_\alpha&&\downarrow_\beta&&\\
H'&\hookrightarrow&G'&\twoheadrightarrow&G'/H'
\end{array}
$$
Then there is a homomorphism $\gamma:G/H\to G'/H'$ such that
$$
\begin{array}{ccccc}
H&\hookrightarrow&G&\twoheadrightarrow &G/H\\
\downarrow_\alpha&&\downarrow_\beta&&\downarrow_\gamma\\
H'&\hookrightarrow&G'&\twoheadrightarrow&G'/H'
\end{array}
$$
commutes. 
proof: It is easy to check that $\gamma(gH)=\beta(g)H'$ is well-defined and gives the desired result.
Proposition: If $\alpha$ is surjective and $\beta$ is injective, then $\gamma$ is injective. 
proof: Suppose $gH\in\ker\gamma$, so that $\gamma(gH)=\beta(g)H'=H'$. We need to show $gH=H$. Well, $\beta(g)\in H'$ and so $\beta(g)=\alpha(h)$ for some $h\in H$ since $\alpha$ is surjective. But, the diagram commutes, so $\alpha(h)=\beta(h)$. Thus, $g=h\in H$ because $\beta(g)=\beta(h)$ and $\beta$ is injective. We now conclude that $gH=hH=H$.
Proposition: If $\beta$ is surjective, so is $\gamma$. In particular, if both $\alpha$ and $\beta$ are isomorphisms, then so is $\gamma$.
proof: pretty obvious.
remark: I stated this result in terms of groups, but this is still true in larger generality. In particular, it works in the category of $G$-modules (or $R$-modules), where $\alpha$ and $\beta$ should be module maps.
Now, in the proof in your question, you have $\mu_n$ and $\mu_{n-1}$ are isomorphisms, and the following diagram commutes:
$$
\begin{array}{ccccc}
\mathrm{im}(\delta_{n-1})&\hookrightarrow&\ker(\delta_n)&\twoheadrightarrow &\ker(\delta_n)/\mathrm{im}(\delta_{n-1})\\
\downarrow_{\mu_{n-1}}&&\downarrow_{\mu_n}&&\\
\mathrm{im}(\tilde{\delta}_{n-1}) &\hookrightarrow&\ker(\tilde{\delta}_n)&\twoheadrightarrow&\ker(\tilde{\delta}_n)/\mathrm{im}(\tilde{\delta}_{n-1})
\end{array}
$$
