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 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 map taking $([\alpha],[\beta])$ to $[\gamma]$ is just the isomorphism between the limit of this spectral sequence and the total cohomology $H^*(C_{33})$.