Let $s \in [0,\infty)$ and consider the Hilbert space $H^{s}:=W^{2,s}$. Its dual is usually denoted as $H^{-s}$.
Also, let $\langle, \rangle_s$ be the inner product of $H^2$. Then, according to the general theory of Hilbert space, there exists a unique $y \in H^{s}$ for each $T \in H^{-s}$ such that \begin{equation} T(x)=\langle x,y \rangle_s \end{equation} for all $x \in H^s$. Thus, there exists a isometric conjugate-linear surjection from $H^s$ to $H^{-s}$.
However, it is also known that $H^{s} \subset L^2 \subset H^{-s}$ and this inclusion is known to be "proper".
How should I understand these two seemingly contradictory facts? I am quite confused..
Well, all separable Hilbert spaces are isomorphic to $\ell^2(\mathbb{N})$, so there is no contradiction in one Hilbert space being a proper subset of another yet being isomorphic to it.
To see how the inclusions $H^s = H^{-s}$ implied by Hilbert space theory and the natural inclusion $H^s \subset H^{-s}$ are different, compute each one. Let $y \in H^s$. Note that the natural inclusion from $H^s$ to $H^{-s}$ is defined (up to a complex conjugate on $y$ depending on the situation) by $$\langle x, y \rangle_{H^s, H^{-s}} = (x, y)_{L^2}.$$ The image of $y$ under the Hilbert space inclusion, call it $T \in H^{-s}$, satsifies $$\langle x, T \rangle_{H^{s}, H^{-s}} = (x, y)_{H^s} = (\langle \xi \rangle^{s}\hat{x}, \langle \xi \rangle ^{s}\hat{y})_{L^2} = (x, \mathcal{F}^{-1}(\langle \xi \rangle ^{2s}\hat{y}))_{L^2}.$$ Hence the Hilbert space isomorphism is $y \mapsto T = \mathcal{F}^{-1}(\langle \xi \rangle ^{2s}\hat{y})$. The natural inclusion just maps $y$ to $y$, whereas the isomorphism from $H^s$ to $H^{-s}$ maps $y$ to $\mathcal{F}^{-1}(\langle \xi \rangle ^{2s}\hat{y})$, which is a distribution in $H^{-s}$.