$\require{AMScd}$
Let $\Gamma=\{1,\gamma\}$ be a group of order 2. 
In my problem from Galois cohomology of real reductive groups I came to a commutative diagram of $\Gamma$-modules 
(abelian groups with $\Gamma$-action)
\begin{equation*}%\label{e:cd}
\begin{CD}
1 @>>>Q_1 @>>>Q_2 @>>>  Q_3     @>>>  1 \\
@.  @VV{\rho_1}V @VV{\rho_2}V   @VV{\rho_3}V \\
1 @>>>  X_1 @>>> X_2 @>>>   X_3 @>>>  1 \\
@.  @VV{\alpha_1}V @VV{\alpha_2}V@VV{\alpha_3}V \\
1 @>>>  P_1 @>>> P_2 @>>>    P_3 @>>>  1 \\
\end{CD}
\end{equation*}
in which the rows are exact, but *not* the columns 
(and $\alpha_k\circ\rho_k\neq 0$).
The top and bottom rows of the diagram split canonically:
$$Q_2=Q_1\oplus Q_3\quad\text{ and }\quad P_2=P_1\oplus P_3.$$
I consider the *Tate* hypercohomology groups 
$${\Bbb H}^0(\Gamma, Q_3\overset{\rho_3}\longrightarrow X _3)\quad\text{ and }
           \quad{\Bbb H}^0(\Gamma,X _1\overset{\alpha_1}\longrightarrow P_1),$$
where both short complexes are in degrees $(-1,0)$.

Below I construct "by hand" a canonical coboundary homomorphism
$$\delta\colon\, {\Bbb H}^0(\Gamma, Q_3\to X _3)\,\longrightarrow\, 
{\Bbb H}^0(\Gamma,X _1\to P_1),$$

>  **Question.**  *How can I get this coboundary homomorphism from a kind of general theory?*

**Remark.** For a group $\Gamma$ of order 2 (and also for any *cyclic* group $\Gamma$) the Tate cohomology and hypercohomology are periodic with period 2. 
Therefore, our $\delta$ is a map
$${\Bbb H}^1(\Gamma,\, Q_3\to X_3\to 0)\, \longrightarrow \, 
{\Bbb H}^2(\Gamma,\, 0\to X_1\to P_1),$$
where both complexes are in degrees $(-2,-1,0)$.

**Construction.**
We start with $[ q_3,  x_3]\in {\Bbb H}^0(\Gamma, Q_3\overset{\rho_3}\longrightarrow X _3)$.
Here $( q_3,  x_3)\in  Z^0(\Gamma,Q_3\to X _3)$, that is,
\begin{equation}
q_3\in Q_3,\quad x_3\in X_3,\quad
\,^{\gamma\kern -0.8pt} q_3+ q_3=0,\qquad \,^{\gamma\kern -0.8pt} x_3- x_3=\rho_3( q_3).\tag{$*$}
\end{equation}
We lift *canonically* $ q_3$ to
$$ q_2=(0, q_3)\in Q_1\oplus Q_3= Q_2,$$
and we lift $ x_3$ to *some* $ x_2\in X _2$.
We write
$$\alpha_2( x_2)=( p_1, p_3)\in P_1\oplus P_3=P_2,$$
where $ p_3=\alpha_3( x_3)\in P_3$ and $ p_1\in P_1$.
We set
$$ x_1=\,^{\gamma\kern -0.8pt} x_2- x_2-\rho_2( q_2).$$
Since by $(*)$  we have
$$\,^{\gamma\kern -0.8pt} x_3- x_3=\rho_3( q_3),$$
we see that  $ x_1\in X _1$.
We compute:
$$\,^{\gamma\kern -0.8pt} x_1+ x_1=\,^{\gamma\kern -0.8pt}(\,^{\gamma\kern -0.8pt} x_2- x_2)-{}^{\gamma\kern -0.8pt}\rho_2(0, q_3)+
    (\,^{\gamma\kern -0.8pt} x_2- x_2)-\rho_2(0, q_2)=-\rho_2(0,\,^{\gamma\kern -0.8pt} q_3+ q_3)=0$$
by $(*)$. Furthermore,
$$\alpha_1( x_1)=\,^{\gamma\kern -0.8pt}( p_1, p_3)-( p_1, p_3)-( 0,\alpha_3(\rho_3( q_3)))
      =\,^{\gamma\kern -0.8pt} p_1- p_1.$$
Thus $( x_1, p_1)\in Z^0(\Gamma, X _1\overset{\alpha_1}\longrightarrow P_1)$.
We set
$$\delta[ q_3, x_3]=[ x_1, p_1]\in {\Bbb H}^0(\Gamma,X _1\to P_1).$$
A straightforward check shows that the map $\delta$ is a well-defined homomorphism.