$\require{AMScd}$
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:
$$Q_2=Q_1\oplus Q_3\quad\text{ and }\quad P_2=P_1\oplus P_3,$$
and these splittings are compatible:
$$
\alpha_2(\rho_2(0,q_3))=
\big(\,0,\,\alpha_3(\rho_3(q_3))\,\big)\tag{$*$}
$$
for $q_3\in Q_3$.
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,
\begin{align*}
\alpha_1( x_1)&=\,^{\gamma\kern -0.8pt}\alpha_2(x_2)-\alpha_2(x_2)-\alpha_2(\rho_2(q_2))\\
&=\,^{\gamma\kern -0.8pt}( p_1, p_3)-( p_1, p_3)-( 0,\alpha_3(\rho_3( q_3)))\\
&=\big(\,^{\gamma\kern -0.8pt}p_1-p_1,\,^{\gamma\kern -0.8pt}p_3-p_3-\alpha_3(\rho_3(q_3))\big)\\
&=\big(\,^{\gamma\kern -0.8pt}p_1-p_1,\,\alpha_3(\,^{\gamma\kern -0.8pt}x_3-x_3-\rho_3(q_3))\big)\\
&=(\,^{\gamma\kern -0.8pt} p_1- p_1,0)
\end{align*}
by $(*)$ and $(**)$.
Thus
$$\alpha_1(x_1)=\,^{\gamma\kern -0.8pt} p_1-p_1.$$
We see that $(x_1, p_1)\in Z^0(\Gamma, X _1\overset{\alpha_1}\longrightarrow P_1)$.
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.