I get
$$\hat f(w) = \int_{-\infty}^{+\infty}\max(t-1,0)e^{-i\omega t}dt$$ $$= -\int_{-\infty}^{1}(t-1)e^{-i\omega t}dt$$ $$ = \lim_{p\to\infty}\left(\int_{-p}^1e^{-i\omega t} - \int_{-p}^1 t\cdot e^{-i\omega t}\right)$$ $$ = \lim_{p\to\infty}\left[\frac{e^{-i\omega p}}{-i\omega p} - \frac{e^{-i\omega t}(i\omega p -1)}{-\omega^2}\right]^1_{-p}$$
But then I get stuck with limits of imaginary numbers...?
For your problem, the Fourier transform in terms of real variables does not exist. So we should understand that $\omega \in \mathbb{C}$.
We want to find $$\hat{f}(\omega) = \int_{-\infty}^{+\infty}\!dt\,\max(t-1,0) e^{-i \omega t} = \int_{0}^\infty\!dt(t-1) e^{-i \omega t}.$$
The integral only converges in the half-plane with $\mathop{\rm Im}\omega <0$. So assuming this, we obtain $$\hat f(\omega) = \int_1^\infty (t-1) e^{-i\omega t} = \frac{(1+i t \omega) e^{- i \omega t}}{\omega^2} + \frac{e^{- i\omega t}}{i\omega}\biggl|_{t=1}^\infty.$$
Now as $\lim_{t\to\infty} p(t) e^{-i \omega t} =0$ for any polynomial $p(t)$, the contribution from the upper limit vanishes and we obtain the result $$ \hat f(\omega) = -\frac{e^{-i \omega}}{\omega^2}.$$
For the back-transform, note that the contour has to be chosen such as to keep $\hat f(\omega)$ within the region of definition, i.e., $\mathop{\rm Im}\omega <0$.