$$f(t) = \begin{cases} t , & 0 \leq t < 1 \\ 2 - t, & t \geq 1 \end{cases}$$ then $f(t) = t + (2-2t)\mathscr{U}(t-1)$ so $\mathcal{L}\{f(t)\} = \mathcal{L}\{t\} + \mathcal{L}\{(2-2t)\mathscr{U}(t-1)\}$
I got $\displaystyle \frac{1}{s^2} + \left(\frac{2}{s} - \frac{2}{s^2}\right)e^{-s}$ which would be right if not for the $\dfrac{2}{s}$ term but I don't know how it's not supposed to be there.