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