= Solution
Orient the alpha curves and choose generators $x,y$ dual to $\alpha_1,\alpha_2$, respectively. Following $\beta_1$ once around the edge identifications and through the attached annulus gives, up to cyclic permutation and simultaneous inversion conventions,
$$
r=x^2y^{-1}x^{-1}yx\,yx^{-1}y^{-1}x.
$$
The dotted blue arcs pass underneath the attached annulus, so their apparent crossings with the red arc on the annulus are not intersections. The <fundamental group> presentation is therefore
$$
\boxed{\pi_1(M_{\mathcal H},x_0)
\cong\langle x,y\mid x^2y^{-1}x^{-1}yx\,yx^{-1}y^{-1}x\rangle.}
$$
The exponent sums of $x,y$ in $r$ are $2,0$. Hence
$$
\boxed{H_1(M_{\mathcal H};\mathbb Z)
\cong\mathbb Z\langle[y]\rangle\oplus(\mathbb Z/2)\langle[x]\rangle,}
$$
and abelianization sends $x$ to the order-two generator and $y$ to the infinite cyclic generator.
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