[Corrected: previous version contained error in description of the attaching map.]
$\Delta_{\alpha \beta}$ is the map $\partial D_\alpha^3 \to S_\beta^2$ obtained by composing the following two maps:
the attaching map $\partial D_\alpha^3 \to X^2$ from the boundary of the $\alpha$th 3-cell to the 2-skeleton of the whole space;
the quotient map $X^2 \to S^2_\beta$ collapsing the complement of the $\beta$th 2-cell to a point.
(See page 141 of Hatcher.)
This is exactly what is described in the sentence that you highlighted.
The cube is the only 3-cell. Its boundary is the surface of the cube. The 2-skeleton of the space is also the surface of the cube, except that opposite faces are identified. The attaching map from the boundary of the 3-cell to the 2-skeleton is projection map on the surface of the cube that identifies opposite faces.
There are three 2-cells, indexed by $\beta$. A 2-cell is represented in your diagram as a pair of opposite faces on the cube, which are identified with each other. The complement of this 2-cell consists of the other two pairs of faces. So the quotient map on the 2-skeleton collapses these other two pairs of faces to a point, while preserving the pair of faces that make up our chosen 2-cell. What remains of the 2-skeleton after this collapse is a 2-sphere. (Remember, the two faces in our chosen 2-cell which survive this collapse are really a single face, because they are identified; this is why we get a single 2-sphere after this collapse rather than a wedge product of two 2-spheres.) The interior of our chosen 2-cell maps onto the 2-sphere minus a single point, and this single point is the image of the other two pairs of faces.
So composing these two maps, we see that $\Delta_{\alpha \beta}$ is a map from the surface of the cube to an $S^2$. There exists a point $p \in S^2$ such that $\Delta_{\alpha \beta}$ maps the interiors of each of the two faces making up our 2-cell homeomorphically to $S^2 \setminus p $, and such that $\Delta_{\alpha \beta}$ maps the other four faces to $p$. Viewing the surface of the cube as an $S^2$ too, the map $\Delta_{\alpha \beta}$ can thought of as a map $S^2 \to S^2$.
Added on request: How to explain that $\Delta_{\alpha\beta} : \partial D^3 \cong S^2 \to S^2$ has degree zero.
Fix a generator $[\sigma] \in H^2 (\partial D^3) \cong \mathbb Z.$
To show that $\Delta_{\alpha \beta}$ has degree zero, we must show that $(\Delta_{\alpha \beta})_\star ([\sigma])$ is the zero element
in $H^2 (S^2)$. Let $r$ be the reflection $\partial D^3 \to \partial D^3$ that exchanges the two faces making up the $\beta$th 2-cell.
From our description of $\Delta_{\alpha \beta}$ (the sentence highlighted in yellow in your original question), it is clear that
$\Delta_{\alpha \beta} \circ r = \Delta_{\alpha \beta}$. This implies that
$ (\Delta_{\alpha \beta})_\star (r_\star ([\sigma])) = (\Delta_{\alpha \beta})_\star ([\sigma]).$
But $r_\star([\sigma]) = - [\sigma]$ since $r$ is a reflection on a sphere. So we have $- (\Delta_{\alpha \beta})_\star ([\sigma]) = (\Delta_{\alpha \beta})_\star ([\sigma])$,
hence $(\Delta_{\alpha \beta})_\star ([\sigma]) = 0$. Therefore, $\Delta_{\alpha \beta}$ has degree zero.
