Problem says:
$(GF)^{*}=F^{*}G^{*}$ , that is, for any form $\xi$ on $P$ , $(GF)^{*}(\xi)=F^{*}(G^{*}(\xi))$
So, by hint, I showed 0-form and 1-form cases.
Let $f$ be a function on $P$ . Then $(GF)^{*}(f) = f(GF) = (fG)F = (G^{*}(f))F = F^{*}(G^{*}(f))$
Now, let $\varphi$ be a 1-form on $P$ . Then $F^{*}G^{*}(\varphi)(v)$ $= F^{*}(\varphi(G_{*}(v)))$ $= \varphi(G_{*}F_{*}(v))$ $= \varphi((GF)_{*}(v))$ $= (GF)^{*}(\varphi)(v)$
So, it remains to show that it holds for any 2 form on $M$.
I tried as follows:
$(GF)^{*}(\eta)(v,w)$ $= \eta((GF)_{*}(v),(GF)_{*}(w))$ $= \eta(G_{*}F_{*}(v),G_{*}F_{*}(w))$ $= G^{*}(\eta)(F_{*}(v),F_{*}(w))$ $= G^{*}F^{*}(\eta)(v,w)$
But this is absurd. How can I fix it?
