Computing the homology groups of the twice-punctured disk

Many thanks to Steve D, user17786, and Dave Hartman for their helpful corrections.

First, I put a cell structure on the twice-punctured disk with 3 0-cells, 5 1-cells, and 1 2-cell:

Note that the boundary of the 2-cell $D$ is $$d_2D=\alpha+\beta+\gamma-\beta+\delta+\epsilon-\delta=\alpha+\gamma+\epsilon,$$ and that the boundaries of the 1-cells are $$\begin{align} d_1\alpha&=0\\ d_1\beta&=y-x\\ d_1\gamma&=0\\ d_1\delta&=z-x\\ d_1\epsilon&=0 \end{align} $$ Now, we identify $y$ with $z$, and $\gamma$ with $\epsilon$, to produce a cell structure on $X$:

For $X$, the chain groups are $$\begin{align} C_0(X)&=\langle x,y\rangle\\ C_1(X)&=\langle \alpha,\beta,\gamma,\delta\rangle\\ C_2(X)&=\langle D\rangle \end{align}$$ where $D$ is our 2-cell, and we have $$\begin{align} d_1\alpha&=0\\ d_1\beta&=y-x\\ d_1\gamma&=0\\ d_1\delta&=y-x \end{align} $$

$$d_2D=\alpha+\beta+\gamma-\beta+\delta+\gamma-\delta=\alpha+2\gamma.$$

Thus, $$H_0(X)=\ker(d_0)/\mathrm{im}(d_1)=\langle x,y\rangle/\langle y-x\rangle=\left\langle\overline{x}\right\rangle\cong\mathbb{Z}$$ $$H_1(X)=\ker(d_1)/\mathrm{im}(d_2)=\langle \alpha,\gamma,\beta-\delta\rangle/\langle \alpha+2\gamma\rangle=\left\langle\overline{\gamma},\overline{\beta-\delta}\right\rangle\cong\mathbb{Z}^2$$ $$H_2(X)=\ker(d_2)/\mathrm{im}(d_3)=0/0\cong 0.$$

$X$ as mentioned earlier is a punctured Klein Bottle, hence deformation retracts onto wedge of two circles. So $H_1(X) = \mathbb{Z} \oplus \mathbb{Z}$.