Consider the vector field in real space $$ \mathbf{f}(\mathbf{r}) = \frac{\mathbf{r}}{|\mathbf{r}|}. $$
What is its Fourier transform vector field $$\hat{\mathbf{f}}(\mathbf{k}) = \int \mathbf{f}(\mathbf{r})\,e^{-i\mathbf{k} \cdot \mathbf{r}} \,\mathrm d \mathbf{r}?$$
Since I am more familiar with it, let me work with the convention $$ \mathcal{F}(g)(y) = \widehat{g}(y) = \int_{\Bbb R^d} g(x)\,e^{-2i\pi x\cdot y}\,\mathrm d x. $$ Your convention is then just $\widehat{g}(\frac{k}{2\pi})$. Since $f$ is not in $L^1$ or $L^2$, one has to consider its Fourier transform in the sense of distributions (in particular, the integral notation is purely formal, $\widehat{f}$ is defined as the tempered distribution such that for any test function $\varphi$ in the Schwartz class, $\langle\widehat{f},\varphi\rangle = \langle f,\widehat{\varphi}\rangle$ where $\widehat{\varphi}$ is given by the integral formula).
Let $\omega_d = \frac{2\,\pi^{d/2}}{\Gamma(d/2)}$, which is the measure of the unit sphere of $\Bbb R^d$ when $d$ is an integer. Then in dimension $d\geq 2$ $$\boxed{ \mathcal{F}\!\left(\frac{x}{|x|}\right) = \frac{2}{i\,\omega_{d+1}}\,\mathrm{vp}\!\left(\frac{x}{|x|^{d+1}}\right) }$$ where $T:= \mathrm{vp}\!\left(\frac{x}{|x|^{d+1}}\right)$, the principal value of $\frac{x}{|x|^{d+1}}$, is the distribution such that for any test function smooth and compactly supported $\varphi\in C^\infty_c(\Bbb R^d)$ $$ \langle T,\varphi\rangle = \frac{1}{2}\int_{\Bbb R^d} \tfrac{x}{|x|^{d+1}}\,(\varphi(x)-\varphi(-x))\,\mathrm d x $$ which coincides outside of $x=0$ with the function $\frac{x}{|x|^{d+1}}$. In the case of the dimension $d=1$, the result is well-known since your function is just the sign function, and the above and can be written $$ \mathcal{F}\!\left(\mathrm{sgn}\right) = \frac{1}{i\,\pi}\,\mathrm{vp}\!\left(\frac{1}{x}\right). $$
Proof. Let me take $d\geq 2$ since the result is well-known in dimension $1$ and can be found in standard books or on this forum (for example here). In dimension $d\geq 2$, the following is well-known (see e.g. the Chapter on the Fourier transform in the book Analysis by Lieb and Loss): the Fourier transform in the sense of distributions of the Function $\frac{1}{|x|}$ is given by $$ \mathcal{F}\!\left(\frac{1}{|x|}\right)(y) = \frac{2}{\omega_{d-1}\,|y|^{d-1}}. $$ By the formula $\mathcal{F}(x\,f)(y) = \frac{i}{2\pi} \,\nabla\widehat{f}(y)$ where the gradient is taken in the sense of distributions, we deduce that $$ \mathcal{F}\!\left(\frac{x}{|x|}\right)(y) = \frac{i}{\pi\,\omega_{d-1}}\, \nabla\frac{1}{|y|^{d-1}}. $$ The distribution $\nabla\frac{1}{|x|^{d-1}}$ coincides outside of $x=0$ with the function $\left(1-d\right) \frac{x}{|x|^{d+1}}$. Due to the singularity at $x=0$, this is however not a locally integrable function and one has to compute it more precisely as a distribution. By definition of the distributional gradient and the fact that $\tfrac{1}{|x|^{d-1}}$ is locally integrable, for any test smooth compactly supported test function $\varphi$, $$ \langle \nabla\tfrac{1}{|x|^{d-1}},\varphi\rangle = -\int_{\Bbb R^d} \tfrac{1}{|x|^{d-1}}\,\nabla\varphi(x)\,\mathrm d x. $$ Using the change of variable $x\to -x$ also gives $$ \langle \nabla\tfrac{1}{|x|^{d-1}},\varphi\rangle = \int_{\Bbb R^d} \tfrac{1}{|x|^{d-1}}\,\nabla\varphi(-x)\,\mathrm d x. $$ Summing the two previous equations yields $$ 2\,\langle \nabla\tfrac{1}{|x|^{d-1}},\varphi\rangle = \int_{\Bbb R^d} \tfrac{1}{|x|^{d-1}}\,\nabla(\varphi(-x)-\varphi(x))\,\mathrm d x. $$ The result then follows by integrating by parts and noticing that $$ \frac{i\,(1-d)}{\pi\,\omega_{d-1}} = \frac{(d-1)\,\Gamma((d-1)/2)}{i\,2\,\pi^{(d+1)/2}} = \frac{\Gamma((d+1)/2)}{i\,\pi^{(d+1)/2}} = \frac{2}{i\,\omega_{d+1}}. $$