Before I answer the question as stated, let me point out that you can also prove the claim as follows: the transfer is induced by a stable map $Y_+\rightarrow X_+$, and the pth Adams operation is a stable cohomology operation after inverting p, so the result is just naturality of cohomology operations.

Now for Quillen's argument. Embed $X$ into $S^n$ and so factor the map f as $X\rightarrow Y\times S^n \rightarrow Y$. If the normal bundle of the first map is, say, Spin oriented, then the Gysin map is defined as the composite: $KO(X) \rightarrow KO(Th(\nu)) \rightarrow KO(S^n \times Y) \rightarrow KO(Y)$. Here we use the Thom isomorphism, then the collapse map, then suspension isomorphism- I'm writing on a phone so I didn't put in the proper indices on the KO's or that some of those should be reduced K-theory.

Anyway, in this case the normal bundle in question is always trivial, and the Thom isomorphism for a trivial bundle is just the suspension isomorphism. This is what Quillen means by doing a suspension isomorphism and then undoing it.

The two arguments aren't unrelated of course, especially if you recall how the stable map yielding the transfer is constructed.