This has a simple interpretation in terms of spectral sequences. Think of the top left 2x2 square of the original square as a triple complex. Call the 3 dimensions $x$ (horizontal), $y$ (vertical), and $z$ ($C_{ij}$ differential). By using either double complex spectral sequence, we see that the total cohomology of the $xy$-plane is just $C_{33}$. Thus the total cohomology of the triple complex is $H^*(C_{33})$.
On the other hand, we can also compute the total cohomology of the triple complex by a spectral sequence that first takes the $z$-cohomology and then takes the $xy$-cohomology. A pair $([\alpha],[\beta])$ in your lemma gives a class that survives this second spectral sequence: $g([\alpha])-u([\beta])$ is the $d_1$ differential, and the $d_2$ differential will vanish for degree reasons. The mapoperation taking $([\alpha],[\beta])$ to $[\gamma]$ is just the isomorphism between the limit of this spectral sequence and the total cohomology $H^*(C_{33})$.
Note that in your proof, $\chi$ is only defined up to a cocycle in $C_{22}$, and so $[\gamma]$ will only be defined modulo the image of $vh:H^{k-1}(C_{22})\to H^{k-1}(C_{33})$. This indeterminacy reflects exactly the fact that $([\alpha],[\beta])$ corresponds to an element of the associated graded of a filtration on $H^{k-1}(C_{33})$ (whose first term is the image of $vh$), rather than an element of $H^{k-1}(C_{33})$ itself.