= Solution
A generator selects one point from each alpha circle and each beta circle. Here the points $1,3$ lie on $\alpha_1\cap\beta_2$, $2$ on $\alpha_1\cap\beta_1$, $a,c$ on $\alpha_2\cap\beta_1$, and $b$ on $\alpha_2\cap\beta_2$. Hence, with $R=\mathbb Z[U]$, the <Heegaard Floer chain complex> has free basis
$$
\boxed{(1,a),\ (1,c),\ (3,a),\ (3,c),\ (2,b).}
$$
A pair such as $(2,a)$ is invalid because it uses beta-one twice.
Orient the attaching circles so that their algebraic-intersection matrix is $\left(\begin{smallmatrix}1&-2\\-2&1\end{smallmatrix}\right)$. The presentation of $H_1(Y)$ is then $m_1-2m_2=0$, $m_2-2m_1=0$, giving $H_1(Y)=\mathbb Z/3$ with $m_2=-m_1$. The upper rectangular domain connects $(2,b)$ to $(3,c)$; the lower one connects $(2,b)$ to $(1,a)$. Thus these three generators have zero joining-cycle obstruction. Changing just the point on the second alpha circle from $a$ to $c$, or just the point on the first from $1$ to $3$, gives the two opposite nonzero meridian classes. This distinguishes the other two generators. The equivalence classes of <Whitney disk classes> are therefore
$$
\boxed{\{(1,a),(2,b),(3,c)\},\qquad\{(1,c)\},\qquad\{(3,a)\}.}
$$
The dot in the central unshaded region represents the basepoint. Each displayed rectangle has $n_z=0$, Euler measure zero and four corner averages $1/4$, hence <Maslov index> one. The <relative grading in Heegaard Floer homology> satisfies
$$
\operatorname{gr}(2,b)-\operatorname{gr}(1,a)=\operatorname{gr}(2,b)-\operatorname{gr}(3,c)=1.
$$
Thus a convenient normalization is \b[degree zero for $(1,a),(3,c)$ and degree one for $(2,b)$]; either singleton can independently be put in degree zero. As before, multiplication by $U^m$ subtracts $2m$ from every degree.
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