Let $H^s(\mathbb{R}^n)$ be the Sobolev space of order $s>1$.
I want to show that
the norms
$\|f\|_s^2:=\int\limits_{\mathbb{R}^d} (1+|\xi|^2)^s |\mathscr{F}f(\xi)|^2d\xi$
and
$[f]_s^2:= \|f\|_{H^{m}(\mathbb{R}^n)}^2+\int\limits_{\mathbb{R}^d\times \mathbb{R}^d} \frac{|f(x)-f(y)|^2}{|x-y|^{n+\sigma 2}}d(x,y)$
are equivalent on that space, where $m$ is the smallest integer such that $\sigma:=s-m\in (0,1)$
I know that for smooth functions $f$ one has $\mathscr{F}(\partial^m f)=i^\alpha x^\alpha \mathscr{F}f$ but I still stuck somehow, when doing the calculations.
So, first, there is an error in your formula, it should be $$ [f]_s^2 = \|f\|_{H^m}^2 + \iint \frac{|\nabla^mf(x)-\nabla^mf(y)|^2}{|x-y|^{d+2\sigma}}\,\mathrm d x \,\mathrm d y. $$ Define the Fourier transform $$ \widehat{u}(y) = \mathcal F(u)(y) = \int_{\mathbb R} e^{-2iπ \,x\cdot y} \,u(x)\,\mathrm d x. $$ Recall that for $\sigma\in(0,1)$, the fractional Laplacian verifies $\mathcal{F}((-\Delta)^{\sigma/2} u) = |2\pi y|^\sigma \,\widehat{u}(y)$ and also $$ (-\Delta)^{\sigma/2} u(x) = c_{d,2\sigma} \int\frac{u(x)-u(y)}{|x-y|^{d+\sigma}} \,\mathrm d y $$ which is just that fact that the Fourier transform of a product is the convolution of the Fourier transforms (see e.g. the last part of the answer here). From the integral formula and by symmetry in $x$ and $y$ it follows that $$ \int u\,(-\Delta)^{\sigma} u = \frac{c_{d,s}}{2} \iint \frac{|u(x)-u(y)|^2}{|x-y|^{d+2\sigma}}\,\mathrm d x \,\mathrm d y =: |u|_{\dot{H}^\sigma}^2. $$ On the other hand, by Plancherel formula $$ \int u\,(-\Delta)^{\sigma} u = \int |2\pi y|^{2\sigma}\, |\widehat{u}(y)|^2\,\mathrm d y. $$ Therefore, taking $u = \nabla^mf$, $$ |\nabla^mf|_{\dot{H}^\sigma}^2 = (2\pi)^{2s} \int |y|^{2\sigma}\, |y^{\otimes m}\widehat{u}(y)|^2\,\mathrm d y \\ = (2\pi)^{2s} \int |y|^{2s}\, |\widehat{u}(y)|^2\,\mathrm d y. $$ On the other hand, by Plancherel again $\|f\|_{H^m}^2 = \|f\|_{L^2}^2 + \|\nabla^m f\|_{L^2}^2 = \int (1+|2\pi y|^{2m})\, |\widehat{f}|^2\,\mathrm d y$, hence finally $$ [f]_s^2 = \int (1+|2\pi y|^{2m}+\tfrac{2}{c_{d,2\sigma}}|2\pi y|^{2s})\, |\widehat{f}(y)|^2\,\mathrm d y. $$ The equivalence between your two norms now just follow from the fact that the function $$ y \mapsto \frac{1+|2\pi y|^{2m}+\tfrac{2}{c_{d,2\sigma}}|2\pi y|^{2s}}{(1+|2\pi y|^2)^s} $$ is bounded above and below by positive constants.