I think the situation is maybe more clear in the case $(H^1)^*$ and $H^1$, which are also isomorphic Hilbert spaces by the Riesz isomorphism w.r.t. the $H^1$ scalar product.
In short
The problem is that the following diagram does not commute!
$$
\begin{array}{ccccccc}
H^1 & \hookrightarrow & L^2 \\
\downarrow & \unicode{x21af} & \downarrow \\
(H^1)^* & \hookleftarrow & (L^{2})^*
\end{array} $$
If we use multiple Riesz isomorphisms, it is ambiguous to write $f \in (H^1)^*$ for a function $f \in H^{1}$. Since it is not clear which path we took in the diagram above.
Therefore we should stick to one isomorphism and not use multiple (non-commuting) ones!
An example
We consider $\Omega = [0,1]$ and $H^1([0,1])$. This space contains a function defined via $f(x) = x$.
- using the embedding into $L^2$, we get $f_{L^2}(x) = x$
- using the $L^2$-Riesz isomorphism, we find $f_{(L^{2})^*}[ g ] = \int_0^1 x \cdot g(x) \, \mathrm{d} x $, for $g \in L^2$
- and finally, $f_{(H^1)^*}[g] = \int_0^1 x \cdot g(x)$, for $g \in H^1$
Now, what if we use the $H^1$-Riesz isomorphism directly?
The scalarproduct is given by $\langle f , g \rangle_{H^1} = \int f(x) g(x) + f'(x) g'(x)\, \mathrm d x$
Therefore, for $g \in H^1$, we get
$$ \widetilde{f}_{(H^1)^*}[g] = \int x g(x) + g'(x)\, \mathrm d x.
$$
Hence
$$\widetilde{f}_{(H^1)^*}
\neq f_{(H^1)^*}!
$$
As we see in the example, if we apply the isomorphisms implied by the $L^2$ and the $H^1$ scalar products, we would need to use different symbols for the same function, in order to keep track of the way we took from $H^1$ to $(H^1)^*$.
(Notice: Of course, one should be careful with point evaluations, expecially in $L^2$. But to keep the formulas short, I commited this crime here. Feel free to correct the formulas above into terms like
$f_{(L^2)^*}[g] = \int_{0}^1 f_{L^2} g \, \mathrm d \lambda$ instead.)
