As far as I know, the only textbook reference for this approach, which is Poincare's original approach, is Seifert and Threlfall's text "A textbook of topology". It's available in English translation but the original was in German. Moreover, Seifert and Threlfall's proof isn't as efficient as it could be, since they're working entirely with simplicial homology. It's much more efficient to work with both simplicial homology and CW-cohomology (together with the knowledge that simplicial and CW homology / cohomology are canonically isomorphic via the relation to singular homology / cohomology).
This version of the proof only works in the context where your manifold is triangulable, and here it goes:
The idea of the dual cell decomposition in general goes like this. Let $\Delta_n$ be an $n$-simplex, and let $F$ be a facet of $\Delta_n$, meaning the convex hull of some collection of $\Delta_n$'s vertices.
The dual polyhedral bit corresponding to $F$ is the convex hull of the barycentres of all facets $F'$ of $\Delta_n$ which contain $F$ (including $\Delta_n$ itself). So in a tetrahedron $\Delta_3$, if $F$ was an edge, the dual polyhedral bit would be a quadrilateral that intersects $F$ in a single point.
Given a triangulated manifold $M$, if $F$ is a simplex of the triangulation, the dual cell corresponding to $F$ is the union of all the dual polyhedral bits to $F$ in all the top-dimensional simplices containing $F$. If $M$ is $m$-dimensional and $F$ is $k$-dimensional, a little geometry later and you'll see the dual cell is an $(m-k)$-dimensional cell in a genuine CW-decomposition of $M$. Again, in the 3-manifold case, if $F$ were an edge, the dual cell would be a $2$-cell with a single vertex at its centre, decomposed into squares.
That's the basic idea. From there the proof of Poincare duality is very much "follow your nose". It's a fun chase and I encourage you to try to work it out on your own, rather than looking it up.
Moreover, spend as much time as you can thinking about evaluating a homology class $X$ on the dual of a homology class $Y$ (provided $X$ and $Y$ have complementary dimensions). You'll have to be careful about thinking of the simplicial vs. CW-homology when thinking this through, of course.
For those looking for a 3d model, I recently turned this into a 3d print. The data is available at the link.
Let's focus on the surface $\Sigma_3$ of genus $3$ just to be concrete.
Here's the geometric idea:
The basic idea is that the lattice $\mathbb{Z}^3$ has to act by deck transformations on the covering space. So we have to have three curves on $\Sigma_3$ which "unwind" to the grid pictured on the right in the covering space. From there it's not a stretch to the cut-and-paste operation in the picture.
Formally - without exhibiting explicit formulae - I think of this two ways.
Three disjoint essential simple closed curves on $\Sigma_3$, $\alpha_1,\alpha_2,\alpha_3$, generate a copy of $F_3$, the free group on three generators, in $\pi_1\Sigma_3$. Associated to $F_3$ is a covering $\widehat{\Sigma_3}\to \Sigma_3$, with $F_3$ acting by deck transformations on $\widehat{\Sigma_3}$ so that $\widehat{\Sigma_3}/F_3 = \Sigma_3$. Since $Z(F_3)<F_3$, we have a quotient cover $\widehat{\Sigma_3}/Z(F_3)\to \Sigma_3$, with the deck group $F_3/Z(F_3)\cong \mathbb{Z}^3$. Geometrically, take the "asterisk" fundamental domain (the one you get by cutting along the red curves) and glue it so it looks like the Cayley graph of $F_3$. Then quotient by the action of the commutators of the generators.
The fundamental group $\pi_1(\Sigma_3)$ projects onto the first homology group $H_1(\Sigma_3;\mathbb{Z})$. Associated to $H_1(\Sigma_3;\mathbb{Z})$ is an abelian cover $A(\Sigma_3)\to \Sigma_3$. The images of $\alpha_1,\alpha_2,\alpha_3$ generate a copy of $\mathbb{Z}^3$ in $H_1$, which corresponds to a $\mathbb{Z}^3$-covering space of $\Sigma_3$.
I think there should be a nice picture of a fundamental domain of this cover in $\mathbb{H}^2$ by stringing together right-angled dodecagons, but I don't quite see it.
Best Answer
To close the question, let me write my comment as an answer. $\chi(S^2)=2$, not $3$ as can be seen by computing the alternating sum of the Betti numbers.
Alternatively, one may consider either of the two usual cell decompositions of the sphere - either the "one $0$-cell, one $2$-cell" or "one $0$-cell, one $1$-cell, two $2$-cells" description of the sphere - to see that $\chi(S^2) = 1-0+1 = 1-1+2 = 2$.