Hopf invariant formula

So, if for $x\in \partial D^{2n}$ you take $G(x)=g(x)$, then probably you want to do "the same" map over the whole cone. It follows from the definition of the union topology (or whatever it's called) that this extends to a continuous map. Then, look at the induced map on the pairs $(K,K-D^{2n})$ (for either complex $K$), and here the map is just (up to confusing maps on pairs with maps on their quotients) the suspension of $g$. Note that for both complexes, the inclusion $(K,\emptyset)\rightarrow (K,K-D^{2n})$ is a cohomology isomorphism in degree $2n$, and use the fact that suspension induces an isomorphism $\pi_n(S^n)\rightarrow \pi_{n+1}(S^{n+1})$.

Edit

A diagram might help:

    (Cfg,o)    --->     (Cf,o)
       |                  |
       |                  |
       V                  V
(Cfg,Cfg-D^2n) ---> (Cf, Cf-D^2n)

Here, the o's stand for $\emptyset$. The vertical maps are inclusions of pairs, and the horizontal lines are really all just $G$ (as I defined it). The top arrow induces $\alpha_{Cf} \mapsto \alpha_{Cfg}$, and the vertical maps are cohomology isomorphisms in degree $2n$. On the bottom row, since $H^*(X,A)=\tilde{H}^*(X/A)$, this is basically a map $S^{2n}\rightarrow S^{2n}$, which by construction is exactly the suspension $Sg$ of $g$. Then, $\deg(Sg)=\deg(g)$; that is, $S:\pi_{2n-1}(S^{2n-1})\rightarrow \pi_{2n}(S^{2n})$ is an isomorphism. So this induces $\beta_{Cf}\mapsto \deg(g)\cdot \beta_{Cfg}$. By naturality,

$$ H(fg)\cdot \beta_{Cfg} = \alpha_{Cfg}^2=G^*(\alpha_{Cf}^2)=G^*(H(f)\cdot \beta_{Cf})=H(f)\cdot (\deg(g)\cdot \beta_{Cfg}).$$