Let
- $H$ be a $\mathbb R$-Hilbert space
- $(e_n)_{n\in\mathbb N}$ be an orthonormal basis of $H$
- $f:H\to H$ be Fréchet differentiable and $$f_n:=\langle f,e_n\rangle\;\;\;\text{for }n\in\mathbb N$$
- $\mathfrak L(A,B)$ denote the space of bounded, linear operators between normed $\mathbb R$-vector spaces $A$ and $B$ and $\mathfrak L(A):=\mathfrak L(A,A)$
Let $L_n:={\rm D}f_n(x)$ denote the Fréchet derivative of $f_n$ at $x\in H$ for some $n\in\mathbb N$. Then, $L_n$ is an element of $\mathfrak L(H,\mathbb R)$ $\Rightarrow$ $\exists!v\in H$ with $$L_n=\langle\;\cdot\;,v\rangle\tag 1$$ by Riesz' representation theorem. On the other hand, $$L_nu=\langle\underbrace{{\rm D}f(x)}_{=:\;L}u,e_n\rangle\;\;\;\text{for all }u\in H\;.\tag 2$$ Thus, by definition of the adjoint $L^\ast$, $$v=L^\ast e_n\;.\tag 3$$
Now, the concrete shape of $L^\ast$ is not obvious to me. In particular, $L^\ast$ is defined to be $v$, but $v$ is unknown. So, the question is: Is there some more concrete representation of $L^\ast$?
Note that we can find a concrete representation of $L^\ast$ when $H=\mathbb R^d$ for some $d\in\mathbb N$: In that case we obtain $$L(u)=\sum_{i=1}^du_i\frac{\partial f_n}{\partial x_i}(x)=u\cdot\nabla f_n(x)\;\;\;\text{for all }u\in H$$ and hence $$v=\nabla f_n(x)\;,$$ if $n\in\left\{1,\ldots,d\right\}$.
I'm not entirely sure what you are asking, maybe what I am writing does not give you what you are looking for. But there is essentially no difference from the finite dimensional case.
If you write $v=\sum_i v_i e_i$ then
$$L_n(e_i)=\langle e_i, v\rangle = v_i$$
And you have $v=\sum_i L_n(e_i)\ e_i$, which is the same as your equation $\nabla f_n(x) = v$. That this sum is well defined follows from $L_n$ being bounded linear functional, ie $v$ being in $H$, which is something of a circular argument.
If you denote $\nabla_j f(x):=Df(x)e_j$, then this becomes
$$v=\sum_j \langle \nabla_j f(x), e_n \rangle e_j$$
So $L^*(w)=\sum_j \langle \nabla_j f(x), w \rangle e_j$. I don't know if this is more pleasing.