Each row is a short sequence, not necessarily exact, as it has been tensored on the right by a module that may not be flat. The same holds for each column, though in this case it has been tensored on the left.
Where is the matrix exact? Each row and column is the result of a tensor product, hence the third and fourth modules remain exact. Beyond this, F0 and G0 are free, and flat, preserving exactness. The only points that might not be exact are F1×B horizontally, A×G1 vertically, and F1×G1 in both directions.
Next show the diagram commutes. This is a non-issue when mapping into or out of 0, so there are only four squares to consider. Start with F1×G1 and consider the two paths to F0×G0. In each case x,y becomes x,y. When mapping F0×G0 onto A×B, x,y becomes the image of x in A cross the image of y in B. Apply similar reasoning to the other two squares, and the diagram commutes.
Now we're ready to map tor(A,B) into tor(B,A). If you want to be technical, tor(B,A) is not present in our matrix. That module is found by tensoring with A on the right, not the left. However, tensor product is commutative, so the fourth column is isomorphic to the modules that build tor(B,A). And since tor actually operates on isomorphism classes of modules, we can say that the vertical homology of A×G1 is indeed tor(B,A).
Let x be a member of the horizontal kernel of F0×B, representing an element in tor(A,B). Let x have a preimage w above, which leads to w heading right. I use the same letter since arrows mapping into the middle module are injective. Now w is in the vertical kernel by commutativity, and has some preimage w above. Carry this right to get y, a member of tor(B,A). It does not matter which w in the preimage of x was selected, for if w leads to 0, we may go up, right, and right to show y = 0. Verify that this map is a module homomorphism, hence tor(A,B) maps into tor(B,A).
This symbolic argument can be reversed to give a well defined map from tor(B,A) back to tor(A,B). Furthermore, the composition in either direction yields the identity map. Therefore the modules are isomorphic, and tor is commutative.
We will see, on the next page, that ext is not commutative.