It's not always possible to visualize things perfectly! But in this case, the spaces involved are quite reasonable. A useful fact here is that a CW pair $(X,A)$ yields a quotient space $X/A$ which inherits a cell structure from $X$ (p.8 in Hatcher). The cells of $X/A$ are the cells of $X -A$ plus one new $0$-cell, the image of $A$ in $X/A$. You get the attaching maps by composing the original attaching maps with the quotient map.
So to get a cell structure on $S^2/A$, where $A$ consists of the north and south poles, we may choose to view $S^2$ as a CW-complex with $0$-skeleton $A$, two distinct arcs connecting those two points, and two discs attached to the resulting circle. In other words, the $0$-skeleton sits inside the $1$-skeleton as the equator, and the $1$-skeleton sits inside $S^2$ as the equator (and this equator passes through the north and south poles).
With this setup, $S^2/A$ has one $0$-cell corresponding to $A$, two $1$-cells and two $2$-cells. The $1$-skeleton is a wedge of two circles, and each disk is attached to this space via a loop $\gamma$ that encircles both circles of the wedge exactly once. Moreover, for the computation of the $\pi_1$ we may choose the basepoint to be the point where the circles meet, i.e., the $0$-cell. By Proposition 1.26 in Hatcher, we have $\pi_1(S^2/A) = \pi_1(S^1 \vee S^1)/N$, where $N$ is the normal subgroup generated by the loop $\gamma$. But we know that $\pi_1(S^1 \vee S^1) \cong \mathbb{Z} * \mathbb{Z} \cong \langle a,b \rangle$. Therefore $\pi_1(S^2/A) \cong \langle a,b \mid ab\rangle \cong \mathbb{Z}$.