Motivation for the norm in $W^{s,p}(\Omega )$ Let $\Omega \subset \mathbb R^n$ a bounded domain with good condition. In the fractional Sobolev space, 
$$W^{s,p}(\Omega )=\left\{u\in L^p(\Omega )\;\Bigg|\; \frac{|u(x)-u(y)|}{|x-y|^{\frac{n}{p}+s}}\in L^p(\Omega \times \Omega )\right\},$$
and we give to this space the norm
$$\|u\|_{W^{1,s}(\Omega )}=\left(\|u\|_{L^p(\Omega )}^p+[u]_{W^{s,p}(\Omega )}^p\right)^{1/p},$$
where $$[u]_{W^{s,p}(\Omega )}=\left(\iint_{\Omega \times \Omega }\frac{|u(x)-u(y)|^p}{|x-y|^{n+sp}}dxdy\right)^{1/p}.$$

Question
  What is the motivation for $[u]_{W^{s,p}(\Omega )}$ ? Why such an expression ? And if $s=1$ (or maybe $s\to 1$), do we have that $$\lim_{s\to 1}\ [u]_{W^{s,p}(\Omega )}^p=\sum_{i=1}^n\left\|\frac{\partial u}{\partial x_i}\right\|^p_{L^p(\Omega )}\ \ ?$$
  If not, I don't understand where would come from $[u]_{W^{s,p}(\Omega )}$.

 A: Originally, the  space $W^{s,p}(\Omega)$   as you defined is a real interpolation space  between $L^p(\Omega)$ and $W^{1,p}(\Omega)$ 
this usually holds true whenever $\Omega $ is an extension domain.
for more details check this valuable lecture note at page 12 Example 1.1.8 therein you find how exactly $W^{s,p}(\Omega)$ can be constructed from $L^p(\Omega)$ and $W^{1,p}(\Omega)$ via real interpolation method.

For the second question the answer is  NO and Yes 

NO Because in fact for any non-trivial smooth function one always has 
$$\lim_{s\to1} [u]_{W^{s,p}(\Omega )} =\infty$$
But Yes there is a convergence: In other to rectify this anomalous one the factor $(1-s)^{1/p}$ to annihilate the singularity at $s=1$ namely we have the following result from Bourgain-Brezis-Mironescu
$$\color{red}{\lim_{s\to1} (1-s)\iint_{\Omega \times \Omega }\frac{|u(x)-u(y)|^p}{|x-y|^{n+sp}}dxdy = K\left\| \nabla u\right\|^p_{L^p(\Omega )} \sim\sum_{i=1}^n\left\|\frac{\partial u}{\partial x_i}\right\|^p_{L^p(\Omega )}}$$
In addition to your interest we have the following from  Maz'ya, and Shaposhnikov we have
$$\color{red}{\lim_{s\to0} s\iint_{\Bbb R^n \times \Bbb R^n }\frac{|u(x)-u(y)|}{|x-y|^{n+so}}dxdy = K\left\|u\right\|^p_{L^p(\Bbb R^n )} }$$
