$W^{s,p}$ is the inhomogeneous Triebel-Lizorkin space $F_{p,q}^{s}(\mathbb{R}^{n})$, with $q=2$, defined by
$$\|f\|_{F_{p,q}^{s}}=\left\|\left(\sum_{k}2^{kqs}|P_{k}f|^{q}\right)^{1/q}\right\|_{L^{p}}$$
As you point out, one obtains the Besov space $B_{p,q}^{s}$ simply by interchanging the order in which norms are taken. I think you would agree that interchanging norms is, in general, a nontrivial action. It is clear from Minkowski's integral inequality that
$$\|f\|_{F_{p,q}^{s}}\leq\|f\|_{B_{p,q}^{s}} \enspace p\geq q, \quad\|f\|_{B_{p,q}^{s}}\leq\|f\|_{F_{p,q}^{s}} \enspace q\geq p$$
Additionally, by the nesting property of sequence spaces,
$$\|f\|_{F_{p,q}^{s}}\leq\|f\|_{B_{p,r}^{s}} \enspace q\geq r, \quad \|f\|_{B_{p,r}^{s}}\leq\|f\|_{F_{p,q}^{s}} \enspace r\geq q$$
I believe that an equivalent characterization of $B_{p,q}^{s}$, for $0<s<1$, is in terms of the norm
$$\|f\|_{L^{p}}+\left(\int_{\mathbb{R}^{n}}\dfrac{(\|f(x+t)-f(x)\|_{L^{p}})^{q}}{|t|^{n+s}}dt\right)^{1/q}$$
If this is correct, then Besov spaces correspond to the generalized Lipschitz spaces $\Lambda_{\alpha}^{p,q}$ in E.M. Stein, Singular Integrals and Differentiability Properties of Functions, Chapter 5, where $s=\alpha$ in our notation. Furthermore, one can show using this characterization that
$$W^{s,p}(\mathbb{R})\not\subset B_{p,q}^{s}(\mathbb{R}), \quad q<2 \tag{1}$$
and
$$B_{p,q}^{s}(\mathbb{R})\not\subset W^{s,p}(\mathbb{R}), \quad q>2 \tag{2}$$
According to section 6.8 of the aforementioned reference, the function
$$f_{s,\sigma}(x):=e^{-\pi x^{2}}\sum_{k=1}^{\infty}a^{-ks}k^{-\sigma}e^{2\pi i a^{k}x}, \quad x\in\mathbb{R}$$
where $a>1$ is an integer, satisfy
$$f_{s,\sigma}\in W^{s,p}(\mathbb{R})\Leftrightarrow \sigma>\dfrac{1}{2},\quad \forall 1<p<\infty$$
and
$$f_{s,\sigma}\in B_{p,q}^{s}(\mathbb{R})\Leftrightarrow \sigma>\dfrac{1}{q},\quad\forall 1<p<\infty$$
From this result, which I imagine depends on results for lacunary Fourier series, it is easy to deduce (1) and (2).
Answer of $(2)$ In this answer, I will use $L^2,L^\infty$ for simplicity instead of $L^2\big([0,1],\Bbb K\big)$ and $L^\infty\big([0,1],\Bbb K\big)$. Also, these are sets of equivalence classes, but I will not distinguish in between equivalence classes and their representatives, as any two representatives are almost everywhere equals.
From the above, we have $$\int|\alpha\cdot f|^2\leq||\alpha||_\infty^2\int|f|^2,\text{ hence }||M_\alpha||\leq||\alpha||_\infty.$$ To prove the reverse direction, assume at first $\alpha$ is simple function with $\alpha\not\equiv 0$ and consider the measurable set $E=\alpha^{-1}(c)$, where $|c|=||\alpha||_\infty$. Since, $\alpha$ is non-trivial simple function we have $m(E)>0$, as $|c|$ is the maximum value of the simple function $|\alpha|$. Now, letting $$f=\frac{\chi_E}{\sqrt{m(E)}}\text{ we have }||f||_2=1 \text{ and }$$$$||M_\alpha(f)||_2=\frac{1}{\sqrt{m(E)}}\bigg(\int_E|\alpha|^2\bigg)^{1/2}=\frac{1}{\sqrt{m(E)}}\bigg(m(E)||\alpha||_\infty^2\bigg)^{1/2}=||\alpha||_\infty.$$
So, we have $||M_\alpha||=||\alpha||_\infty$, in this case. But, $\alpha\equiv 0$ implies $||M_\alpha||=0=||\alpha||_\infty$, trivially. So, we are done for simple functions.
Now, consider arbitrary essentially bounded function $\psi$, and since simple functions are dense in $L^\infty$ we have a sequence of simple functions $\{\varphi_n\}$ with $\lim||\psi-\varphi_n||_\infty=0$. Now, $$||M_\psi f-M_{\varphi_n}f||_2=||M_{\psi-\varphi_n}f||_2\leq ||\psi-\varphi_n||_\infty||f||_2\text{ for all }f\in L^2$$$$\implies \lim||M_\psi -M_{\varphi_n}||\leq\lim||\psi-\varphi_n||_\infty=0 $$$$\implies ||M_\psi||=\lim ||M_{\varphi_n}||=\lim||\varphi_n||_\infty=||\psi||_\infty.$$
So, we are done.
Best Answer
So, here I consider $1<p<\infty$ and write $\frac{1}{p}+\frac{1}{q}=1$ for some $1<q<\infty$.
Next let $F\in \ell_p'$ i.e. $F$ is a bounded or continuous real linear functional on $\ell_p$. Set, $$x_n:=\sum_{j=1}^n\big|F(e_j)\big|^{q-1} \operatorname{sgn}\big(F(e_j)\big)e_j.$$ Here, as usal $e_j$ is the sequence having $j$-th term as $1$ and all other terms of $e_j$ are zero. Also, for any real $r\in \Bbb R$ we define $$\operatorname{sgn}(r):=\begin{cases} 1&\text{ if }r\geq 0,\\ -1&\text{ if }r<0.\end{cases}$$ Now, $$\sum_{j=1}^n\big|F(e_j)\big|^{q}= F(x_n)\leq \big|F(x_n)\big|\leq \|F\|\cdot \|x_n\|_p=\|F\|\cdot \bigg(\sum_{j=1}^n\big|F(e_j)\big|^q\bigg)^{1/p}$$$$\implies \bigg(\sum_{j=1}^n\big|F(e_j)\big|^q\bigg)^{1/q}\leq \|F\|$$$$\implies \big\{F(e_j)\big\}_{j\geq 1}\in \ell_q\text{ as }n\in \Bbb N\text{ is arbitrary}.$$ Also, $$\ell_p\ni \{y_j\}_{j\geq 1}\overset{F}{\longmapsto}\sum_{j=1}^\infty y_j\cdot F(e_j)\in \Bbb R.$$