Like Mr Yuan suggested, call the first one 'dual' and write $f^\ast$ and the second one adjoint and write $f^\dagger$. Then a fairy simple calculation shows, that $f^\ast$ and $f^\dagger$ are closely related to each other:
Let $i: X \to X^\ast$ and $j: Y \to Y^\ast$ be the operators coming from the Riesz's represantation theorem. Then for any $y' \in Y^\ast$ and $x \in X$ there holds:
$\langle j^{-1}\cdot y', f \cdot x\rangle = \langle f^\dagger \cdot j^{-1} \cdot y, x\rangle$.
On the right hand side we have: $\langle f^\dagger \cdot j^{-1} \cdot y', x\rangle = i \cdot f^\dagger \cdot j^{-1} \cdot y' \cdot x$,
while on the right hand side there is: $\langle j^{-1} \cdot y', f \cdot x \rangle = y' \cdot f\cdot x = f^\ast \cdot y' \cdot x$
That for we get: $f^\ast \cdot y' \cdot x = i \cdot f^\dagger \cdot j^{-1} \cdot y' \cdot x$. Since this holds for all $x \in X$, there must be
$f^\ast \cdot y' = i \cdot f^\dagger \cdot j^{-1} y'$ for all $y' \in Y^\ast$ and we can conclude, that
$f^\ast = i\cdot f^\dagger \cdot j^{-1}$.
If you don't destinguish between $X$ and $X^\ast$ and $Y$ and $Y^\ast$ respectively, then $f^\ast = f^\dagger$.
Kind regards
Konstantin
To the contrary, my feeling is that nonseparable Hilbert spaces are in some sense artifacts and can almost always be avoided. And more generally, to your comment about nonseparable Banach spaces being "nothing special", most of the nonseparable Banach spaces one meets in practice are duals of separable Banach spaces, and therefore are weak* separable. Algebras of almost periodic functions, and their noncommutative analogs, the CCR algebras, are rare examples of interesting nonseparable Banach spaces which are not dual spaces, but even here one is mainly interested in regular representations, which are determined by their behavior on separable subalgebras (the span of the functions $e_s$ with $s$ rational, say).
Best Answer
I have two, and perhaps infinitely many, examples in finite dimension $n$.
n=2. Take $X={\mathbb R}^2$ with $\ell^1$-norm $$\|x\|_1=|x_1|+|x_2|.$$ Then $X^*={\mathbb R}^2$ has the $\ell^\infty$-norm $$\|y\|_\infty=\max(|y_1|,|y_2|).$$ I turns out that $$\|x\|_1=\max(|x_1+x_2|,|x_1-x_2|)$$ and thus $X'$ is isometric to $X$, via $x\mapsto(x_1+x_2,x_1-x_2)$.
More generally, suppose that in $\mathbb R^n$, we have a convex polytope $T$ that is self-dual and is symmetric under $x\leftrightarrow-x$. Let $\|\cdot\|_T$ be the gauge associated with $T$. Then $X=(\mathbb R^n, \|\cdot\|_T)$ is isometric to $X'$ because $T$ is the unit ball of $X$ and $T'=T$ is that of $X'$.
For instance, if n=4, the polyoctahedron (= octaplex) has these properties, thus there is an $\mathbb R^4$ that is isometric to its dual, yet is not Hilbert. If $n\ge3$, the simplex is self-dual but not centro-symmetric.
This raises two questions: