The piecewise function is: $$ \begin{array}{cc} & \begin{array}{cc} t & 0\leq t< 1 \\ 2-t & 1\leq x\leq 2 \\ 0 & t>2 \end{array} \end{array} $$ Now putting the piecewise into LaPlace form: $$\int_0^1te^{-st}dt +2\int_1^2e^{-st}dt-\int_1^2te^{-st}dt+0$$ The first and third integral are the same integration by parts: $u=t,du=dt,dv=e^{-st},v=-\frac{1}{s}e^{-st}$.
Then the expression becomes: $$-\frac{t}{s}e^{-st}+\int_0^1\frac{1}{s}e^{-st}dt+2\int_1^2e^{-st}dt-\frac{t}{s}e^{-st}+\int_1^2\frac{1}{s}e^{-st}dt$$ Which after integrating and evaluating equals: $$-\frac{1}{s}e^{-s}-\frac{1}{s^2}e^{-s}+\frac{1}{s^2}-\frac{2}{s}e^{-2s}+\frac{2}{s}e^{-s}-\frac{2}{s}e^{-s}-\frac{2}{s}e^{-2s}-\frac{1}{s^2}e^{-2s}+\frac{1}{s}e^{-s}+\frac{1}{s^2}e^{-s}$$ Are my calculations correct or did I mess up somewhere?
I believe you've made a few small sign slips. Your starting point is correct, $$\tilde f(s) = \int_0^1 t e^{-st} + \int_1^2 (2-t) e^{-st} \; dt$$ If we're not careful, we'll get far too many terms (like your final answer, which is difficult to figure out whether it's correct), so let's evaluate \begin{align} I(t) =\int t e^{-st} \; dt &= -\frac{t}{s} e^{-st} + \frac{1}{s} \int e^{-st} \; dt \\ &= -e^{-st} \left[\frac{t}{s} + \frac{1}{s^2} \right] \end{align} where we're dropping the constant of integration since we're only considering definite integrals in $\tilde f(s)$.
Then \begin{align} \tilde f(s) &= I(1) - I(0) -\frac{2}{s} \left( e^{-2s} - e^{-s} \right) - I(2) + I(1) \\ &= 2 I(1) - I(0) - I(2) - \frac{2}{s} \left( e^{-2s} - e^{-s} \right) \\ &= -2e^{-s} \left[ \frac{1}{s} + \frac{1}{s^2} \right] + \frac{1}{s^2} +e^{-2s} \left[ \frac{2}{s} + \frac{1}{s^2} \right] - \frac{2}{s} \left(e^{-2s} - e^{-s} \right) \\ &= e^{-s} \left[ -\frac{2}{s} - \frac{2}{s^2} + \frac{2}{s} \right] + \frac{1}{s^2} e^{-2s} + \frac{1}{s^2} \\ &= -\frac{2}{s^2} e^{-s} + \frac{1}{s^2} \left(e^{-2s} + 1 \right) \\ &= \frac{e^{-2s} - 2e^{-s} + 1}{s^2} = \left(\frac{1- e^{-s}}{s} \right)^2 \end{align}