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 the following 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 related result which is both more and less general than the one of James. 

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
$$
\begin{CD}
G_{n+1} @> \pi >> S^n \\
@VVV @VV E V \\
F_{n+1} @>>\lambda > \Omega^{n+2} S^{2n+2}
\end{CD}
$$
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
$$
\begin{CD}
B @> \pi >> S^n \\
@VVV @VV E V \\
F_{n+1} @>>\lambda > \Omega^{n+2} S^{2n+2}
\end{CD}
$$
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 onto $\Sigma B \wedge X$. 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 (in Boardman and Steer), we obtain, after some rewriting, the formula
$$
\lambda(h_F) = (f \wedge g) \circ \Sigma p_{12} \qquad (*)
$$
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 $g$ is the identity map of $S^{n+1}$.
The map $\lambda(h_F)$ is adjoint to the composite
$$
\begin{CD}
B\to G_{n+1} \to F_{n+1} @>\lambda >> \Omega^{n+2} S^{2n+2}\, 
\end{CD}
$$
which is one of the composites of the reformulated claim.

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, the last composition is adjoint to the composition
$$
\begin{CD}
B @>>> G_{n+1} @> \pi >> S^n @> E >> \Omega^{n+2} S^{2n+2}
\end{CD}
$$
which is the other composite of the reformulated claim.
Thus the claim follows from  equation $(\ast)$.


**Comments:** (1). The above argument shows that the diagram of Greg's question (1) commutes after looping. The equation $(\ast)$ isn't generally valid when $B$ (or $X$) isn't a suspension. 

For general $B$, 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.

A potential counterexample to the diagram commuting is to take $n = 2$ and  $B =\Bbb RP^3 = SO(3) \subset O(3)$, since the diagonal $\Bbb RP^3 \to \Bbb RP^3 \wedge \Bbb RP^3$ is stably essential (it's detected by the the cup product).

In fact, Bill Richter (unpublished) shows that if $n > 2$ then Greg's diagram fails to homotopy commute (Richter also explicitly identifies the deviation from the diagram commuting).

(2). The reader might ask why we've worked with the Boardman-Steer Hopf invariant rather than the one of James. The answer is that James' invariant doesn't satisfy a Cartan formula––––there are "partial" Cartan formulae, which were known to Barratt, but this is lost knowledge... 

(3). The Boardman and Steer paper I alluded to is:

 Boardman, J. M.; Steer, B. On Hopf invariants. Comment. Math. Helv. 42 1967 180–221.

See also:

Boardman, J. M.; Steer, B. Axioms for Hopf invariants. Bull. Amer. Math. Soc. 72 1966 992–994.

(4) The result of James about the diagram looping after one suspension isn't stated by James. In effect, James proves the (reformulated) claim in the special case when $B = S^p$ is a sphere (so Greg's diagram will commute on homotopy groups).  The result is given by Corollary 15.9 of the paper:

 James, I. M. On the suspension triad. Ann. of Math. (2) 63 (1956), 191–247.