Commutativity relationship between the Malliavin derivative and the Skorkohod divergence operators. 
On proposition $1.3.1$ of Nualart's book "The Malliavin Calculus and
  Related Topics" it's stated the following. Let $H$ be a real separable
  Hilbert space, and let $W=\{W(h),h\in H\}$ be a Gaussian isonormal
  process.
Let $u$ be a process of the form $$u=\sum_{j=1}^n F_j h_j,$$ where the
  $F_j$ are smooth random variables (in the Malliavin calculus sense) 
  and the $h_j$ are elements on the Hilbert space $H$.
We have the following commutativity relationship between the Malliavin
  derivative and the divergence operators.
$$D^h(\delta(u)):=\langle D\delta(u),h\rangle_H=\langle u,
 h\rangle_H+\delta (D^h u) (\star)$$


My attempt:
Using the integration by parts formula I can see that 
$$\delta(u)= \sum_{i=1}^n F_i W(h_i)-\sum_{i=1}^n \langle DF_i,h_i\rangle_H,$$
hence 
$$D^h(\delta(u))=D^h\left(\sum_{i=1}^n F_i W(h_i)-\sum_{i=1}^n \langle DF_i,h_i\rangle_H\right)$$
The linearity of the Malliavin derivative allows us to write:
$$=\sum_i D^h(F_i W(h_i))-\sum_i D^h\langle DF_i,h_i\rangle_H$$
For the first term we can use the product rule and we obtain 
$$=\langle h,u\rangle_H+ \sum_i (D^hF_i)W(h_i)-\color{red}{D^h\langle DF_i,h_i\rangle_H}.$$
At this point the author writes the red term as:
$$D^h\langle DF_i,h_i\rangle_H=\langle D(D^hF_i),h_i\rangle_H$$
(1) Why does this hold?
The definitions tells me:
$$D^h\langle DF_i,h_i\rangle= \langle D \langle DF_i,h_i\rangle, h\rangle=\langle D(D^{h_i}F_i),h\rangle.$$

(2) The Malliavin derivative is an operator that takes a real-valued random variable and returns us an $H$-valued random variable. Why  on $\star$ we have $D^h u:= \langle Du, h\rangle$? $u$ is not a real-valued random variable, how to take the Malliavin derivative of this "stochastic process"?
 A: Let me start by answering the second point. Nate Eldredge points to the right place in Nualart's book in a comment. Here I will just recreate the detail from there that if $G$ is a smooth $H$-valued random variable i.e.
$$G = \sum_{j=1}^n G_j v_j$$
where the $G_j$ are smooth random variables and $v_j \in H$ then e.g. 
$$DG = \sum_{j=1}^n DG_j \otimes v_j$$
is valued in $H \otimes H$. One then checks in much the same way as the real-valued case that this defines a closable operator etc. 
Now I turn to your first question.
Since $F_j$ is smooth, we can write $F_j = f_j(W(e_1), \dots, W(e_n))$ for $\{e_i: i = 1, \dots, n\}$ orthonormal in $H$ and $f_j$ smooth. Then
\begin{align}
D^h \langle DF_j, h_j \rangle =& D^h \left (\sum_{i=1}^n \partial_if_j(W(e_1),\dots,W(e_n)) \langle e_i, h_j \rangle \right)
\\
=& \sum_{k=1}^n \sum_{i=1}^n \partial_k \partial_i f_j(W(e_1), \dots, W(e_n)) \langle e_i, h_j \rangle \langle e_k, h \rangle
\\
=& \sum_{i=1}^n \sum_{k=1}^n \partial_i\partial_k  f_j(W(e_1), \dots, W(e_n))\langle e_k, h \rangle\langle e_i, h_j \rangle
\\
=& \sum_{i=1}^n \partial_i D^h F_j \langle e_i, h_j \rangle
\\ 
=& \langle DD^h F_j, h_j \rangle
\end{align}
as desired.
