$\newcommand{\bQ}{\mathbb{Q}}\newcommand{\bZ}{\mathbb{Z}}\DeclareMathOperator{\Tor}{Tor}\newcommand{\Tors}{\mathrm{Tors}}$I tried to write up the computation with some level of details, please let me know if anything looks dubious.
Consider the long exact sequence $$\ldots\to H_1(\Gamma, M_2)\to H_1(\Gamma,M_3)\to M_{1,\Gamma}\to M_{2,\Gamma}\to M_{3,\Gamma}\to 0$$ which we will view as an (acyclic) complex and denote its terms by $C^n, n\leq 0$, for brevity, so that $C^0=M_{3,\Gamma},C^{-1}=M_{2,\Gamma}$ etc. Consider the spectral sequence associated with the derived functor of $-\otimes_{\bZ}\bQ/\bZ$ and the 'bête' filtration on the complex $\ldots C^1\to C^0$. I'll use the cohomological grading conventions so that the first page of the spectral sequence looks like $E_{1}^{i,j}=\Tor_{-j}^{\bZ}(C^{-i},\bQ/\bZ)$. Note that for an abelian group $M$ we have $\Tor^{\bZ}_0(M,\bQ/\bZ)=M\otimes\bQ/\bZ,\Tor^{\bZ}_1(M,\bQ/\bZ)=M_{\mathrm{Tors}}$, and $\Tor_{>1}(M,\bQ/\bZ)=0$, hence the $1$st page looks like this, having only $2$ potentially non-zero rows:
$$\begin{matrix}\dots & \color{red}{H_1(\Gamma,M_2)\otimes\bQ/\bZ} & \color{red}{H_1(\Gamma,M_3)\otimes\bQ/\bZ} & M_{1,\Gamma}\otimes\bQ/\bZ & M_{2,\Gamma}\otimes\bQ/\bZ & M_{3,\Gamma}\otimes\bQ/\bZ & \\ \dots & H_1(\Gamma,M_2)_{\mathrm{Tors}} & H_1(\Gamma,M_3)_{\Tors} & M_{1,\Gamma,\Tors} & M_{2,\Gamma,\Tors} & M_{3,\Gamma,\Tors}\end{matrix}$$
The differentials $d_{i,j}:E_1^{i,j}\to E^{i+1,j}_1$ are simply the maps induced by the maps in the above exact sequence. Since we started with an acyclic complex, the spectral sequence must converge to zero.
Now, since $H_i(\Gamma, M)$ is annihilated by $|\Gamma|$ for all $\Gamma$-modules $M$, the two entries that are highlighted in red are in fact zero (and all the hidden entries in the top row are zero likewise). This implies that the only possibly non-trivial differential on the second page is $$E_{2}^{-2,0}=\ker (E_1^{-2,0}\to E_1^{-1,0})=\ker (M_{1,\Gamma}\otimes\bQ/\bZ \xrightarrow{i_*} M_{2,\Gamma}\otimes\bQ/\bZ )\to E_2^{0,-1}=\mathrm{coker} (M_{2,\Gamma,\Tors} \xrightarrow{j_*} M_{3,\Gamma,\Tors})$$
This differential must be an isomorphism, while all other $E_2^{i,j}$ must already be zero as otherwise something would survive to the abutment $E_3^{i,j}=E_{\infty}^{i,j}$ of the spectral sequence. But this is exactly saying that the $6$-term sequence in question is exact, with $\delta$ being defined as the inverse to the differential $E_2^{-2,0}\to E_2^{0,-1}$.
Best Answer
$E= \mathbb F_q ( \sqrt{t}, \sqrt{t^2-1} ) $ over $F =\mathbb F_q(t)$ does the trick if $q$ is congruent to $1$ mod $4$. It suffices to check that at each place where one of the extensions ramifies, the other is split, as this clearly gives a cyclic decomposition group, and unramified places are always cyclic.
The first extension ramifies at $0, \infty$. At $t=0$, the second extension $y^2=t^2-1 $ locally looks like $y^2 = 0-1 = -1$ which is split since $q \equiv 1 \bmod 4$. We can write the second extension as $\left(\frac{y}{t}\right)^2= 1- \frac{1}{t^2}$, which at $t=\infty$ locally looks like $\left(\frac{y}{t}\right)^2 = 1- \frac{1}{\infty^2} = 1- 0 =1$, which is split.
The second extension ramifies at $1,-1$ where the first extension locally looks respectively like $y^2=1$ and $y^2=-1$, which again are split since $q\equiv 1\bmod 4$.