There is not much you have to do: you are giving a vector space with an inner product. It's completion is a Hilbert space. The only issue is to show that the inner product is such.
The definition is fine in the sense that the functions $\{1_{[0,t]}:\ t\in[0,T]\}$ are linearly independent, so you can define $\langle f,g\rangle_H$ for all $f,g\in\mathcal E$. What is not so clear to me is that you get an inner product: mainly, you need to show that $\langle f,f\rangle_H=0$ implies $f=0$. This means that if $f=\sum_{j=1}^n\alpha_j\,1_{[0,t_j]}$,
$$
0=\langle f,f\rangle_H=\frac12\,\sum_{k,j}\alpha_k\alpha_j(t_k^{2H}+t_j^{2H}-|t_k-t_j|^{2h}),
$$
implies $\alpha_1=\cdots=\alpha_n=0$. I don't see an immediate way to show that, and that's the crucial issue to solve in your questions. If you show the above, the rest is straightforward.
There are a couple issues with your reasoning, too. It is not true that $t^{1/2}$ and $t^H$ are equivalent, unless $H=1/2$. If you had $H<1/2$ and $t^H\leq c\,t^{1/2}$, you have $t^{1/2-H} \geq1/c$ for all $t$ close to zero, which is false. Similarly with $H>1/2$.
Also, to test that two norms are equivalent you have to do it for all elements of the space, not just some. So you cannot do it for $1_{[0,t]}$ only, you have to do it for all linear combinations.
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.
Best Answer
You may consider an H-valued random variable as a measurable function from $\Omega$ to $H$, where $H$ is equipped with its Borel $\sigma$-algebra. Then as you say, $L^2(\Omega; H)$ is the set of (equivalence classes of) random variables $f$ such that $E[\|f\|^2] < \infty$.
The inner product is $E[\langle f, g\rangle]$ as you guessed; this is indeed an inner product (easy exercise), and moreover it is complete, so $L^2(\Omega; H)$ is a Hilbert space itself.