In my comment to my first answer, I noted that I didn't address Greg's question (1).  This answer aims to address that question. I am indebted to Bill Richter for explaining this argument to me.

We are trying to understand whether or not the square 
$$
\require{AMScd}
\begin{CD}
O(n+1) @>>> S^n \\
@VVV @VVV \\
F_{n+1} @>> H > \Omega^{n+1} S^{2n+1}
\end{CD}
$$
commutes. I mentioned in my comment that James proved it commutes after a loop.  I will sketch a proof of a more general result. 

Let $\lambda = \Sigma H$ denote the Boardman and Steer Hopf invariant. This is a natural transformation of based function spaces
$$
F_\ast(A,B) \to F_\ast(\Sigma^2 A,\Sigma B\wedge \Sigma B)
$$
given by suspending the James-Hopf invariant.

Let's consider the related problem of whether or not the diagram
$$
\begin{CD}
G_{n+1} @> \pi >> S^n \\
@V\cap VV @VV E V \\
F_{n+1} @>>\lambda > \Omega^{n+2} S^{2n+2}
\end{CD}
$$
commutes. If it does then Greg's question (1) will commute if we replace $H$ by $\lambda$, since the map $O(n+1) \to S^n$ factors through $G_{n+1}$.

**Claim:** The last diagram above homotopy commutes after taking a single loop. Reformulate the claim as follows: let $B = \Sigma A$ be a suspension and let $B\to G_{n+1}$ be any map. Then the claim is equivalent to the assertion that the diagram
$$
\begin{CD}
B @> \pi >> S^n \\
@V\cap VV @VV E V \\
F_{n+1} @>>\lambda > \Omega^{n+2} S^{2n+2}
\end{CD}
$$
commutes (by taking $B = \Sigma \Omega G_{n+1}$).

To verify the reformulation of the claim, we will use the Boardman and Steer Cartan formula. Let $X = S^n$, then the action of $G_{n+1}$ on $X$ induces a map 
$$
F: B \times X \to X\, .
$$
The Hopf construction of $F$ is a map $h_F: \Sigma B\wedge X \to \Sigma X$ whose adjoint is the left vertical (inclusion) map of the square. Thus we need to compute $\lambda(h_F): \Sigma B \wedge \Sigma X \to \Sigma X \wedge \Sigma X$.

Take the suspension of this map $\Sigma F: \Sigma (B \times X) \to \Sigma X$. Then we have a factorization up to homotopy
$$
\begin{CD}
\Sigma (B\times X) @>\Sigma F >> \Sigma X \\
    @V p_1 + p_2 + p_{12}  VV @| \\
\Sigma B \vee \Sigma X \vee \Sigma B\wedge X @>>(f,g,h_F) > \Sigma X
\end{CD}
$$
where $p_1$ is the projection onto $\Sigma B$, $p_2$ the projection onto
$\Sigma X$ and $p_{12}$ is the quotient map. The map $f$ is the suspension of the restriction of $F$
to $B \times \ast$, the map $g$ is the suspension of the restriction of $F$ to $\ast \times X$. 

Since $\Sigma F$ is a suspension, its Hopf invariant $\lambda(\Sigma F)$ is trivial. So, if we take the Hopf invariant of the composite 
$$
\Sigma F = (f,g,h_F) \circ (p_1 + p_2 + p_{12})
$$
and use the composition formula and the cartan formula, we obtain, after some rewriting, the formula
$$
\lambda(h_F) = (f \wedge g) \circ \Sigma p_{12} 
$$
Implicit in this calculation, which I've omitted, is the fact that the reduced diagonal maps of both $B$ and $X$ are trivial since these spaces are suspensions. I refer the reader to Boardman and Steer's paper for the details.

Returning to our original notation, note that the map $f$ is the suspension of composition
$$
\begin{CD}
B @>>> G_{n+1} @>\pi >> S^n
\end{CD}
$$
and the map $\lambda(h_F)$ is adjoint to the composite
$$
\begin{CD}
B\to G_{n+1} \to F_{n+1} @>\lambda >> \Omega^{n+2} S^{2n+2}\, ,
\end{CD}
$$
whereas the map $(f \wedge g) \circ \Sigma p_{12}$ is the composite
$$
\begin{CD}
\Sigma^2 (B \times S^n) @>>> \Sigma^2 (G_{n+1} \times S^n) @>q >> \Sigma^2 G_{n+1} \wedge S^n @>\Sigma \pi \wedge 1 >> 
\Sigma S^n \wedge \Sigma S^n
\end{CD}
$$
where $q$ is the quotient map. By definition last composition is adjoint to the composite
$$
\begin{CD}
B @>>> G_{n+1} @> \pi >> S^n @> E >>> \Omega^{n+2} S^{2n+2}
\end{CD}
$$
and the claim is proved!


**Comment:** The above argument shows that the diagram of Greg's question (1) commutes after looping. The equation we derived
$$
\lambda(h_F) = (f \wedge g) \circ \Sigma p_{12} 
$$
isn't valid when $B$ isn't a suspension. In this instance there is a correction term in the formula involving the reduced diagonal $B\to B \wedge B$. This gives evidence that Greg's diagram will fail to commute before taking loops.