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"?