Is the fundamental group of non orientable surface Fuchsian?

Let M be the connected sum of 3 projective planes. Then is the fundamental group of M a Fuchsian group?

This group is an extension of the two element group by the surface group for a genus 2 surface (which is a Fuchsian group). So it seems very close to Fuchsian.

Recall that the connected component of the isometries of the sphere, plane and hyperbolic plane are $ SO_3, SE_2, PSL_2=SO_{2,1} $ respectively.

The nonorientable surface $ N_0 $ (which is just the projective plane) is given by $$ N_0 \cong SO_3/O_2 $$ where $ SO_2 $ is the connected group of rotations around z axis $$ SO_2= \left \{ \ \begin{bmatrix} a & b & 0 \\ -b & a & 0 \\ 0 & 0 & 1 \end{bmatrix} : a^2+b^2=1 \right \} $$ and $$ r:=\begin{bmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{bmatrix} $$ so that $ <r,SO_2> \cong O_2 $. The orientable double cover is the sphere $ \Sigma_0 $ $$ \Sigma_0 \cong S^2 \cong SO_3/SO_2 $$ Similarly for the the non orientable surface $ N_1 $ (the connected sum of two projective planes, in other words the Klein bottle) we have $$ SE_2= \left \{ \ \begin{bmatrix} a & b & x \\ -b & a & y \\ 0 & 0 & 1 \end{bmatrix} : a^2+b^2=1 \right \} $$ there is a connected group $ V $ of translations up each vertical line $$ V= \left \{ \ \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & y \\ 0 & 0 & 1 \end{bmatrix} : y \in \mathbb{R} \right \} $$ There is also an element that shifts horizontally by one unit $$ b:=\begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} $$ Now if we include the rotation by 180 degrees $$ \tau:=\begin{bmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{bmatrix} $$ then $$ N_1 \cong SE_2/<V,b, \tau> $$ And the orientable double cover is the torus $ \Sigma_1 $ $$ \Sigma_1 \cong T^2 \cong SE_2/<V,b> $$ Is something similar true for $ N_2 $ the connected sum of 3 projective planes? The orientable double cover here is a genus 2 surface $ \Sigma_2 $. Then $$ \Sigma_2 \cong \Gamma \backslash SL_2 /SO_2 $$ where $ \Gamma $ is a Fuchsian surface group. Is there another Fuchsian group $ \Gamma' $ containing $ \Gamma $ as a index two subgroup such that $$ N_2 \cong \Gamma' \backslash SL_2 /SO_2 $$

Yes, you are misinterpreting what is said in the link.

The proof is by contradiction. Suppose that $\pi=\pi_1(M)$ is isomorphic to a Fuchsian subgroup $\Gamma< G= PSL(2,R)$. The group $G$ acts (and properly) conformally on the upper half-plane $U$. Thus, $\Gamma$ acts properly discontinuously and conformally on $U$ as well. Hence, the quotient $S=U/\Gamma$ is a Riemann surface (a 1-dimensional complex manifold) satisfying $\pi_1(S)\cong \Gamma\cong \pi = \pi_1(M)$. Hence, due to asphericity of $M$ and contractibility of $U$, the surfaces $S$ and $M$ are homotopy-equivalent. Since $M$ is compact, you get $H_2(M, Z_2)=Z_2$. Hence, $H_2(S, Z_2)=Z_2$. From this, you conclude that $S$ is also compact. But $S$ is orientable, hence, $H_2(S)\cong Z$, while $H_2(M)=0$.