Consider the Laplace transform of $\color{green}{t\cfrac{\mathrm{d^2}f}{\mathrm{d}t^2}}$:
$$\mathcal{L}\left[{t\cfrac{\mathrm{d^2}f}{\mathrm{d}t^2}}\right]=\int_{t=0}^{\infty}e^{-st}{t\cfrac{\mathrm{d^2}f}{\mathrm{d}t^2}}\mathrm{d}t=-\cfrac{\mathrm{d}}{\mathrm{d}s}\int_{t=0}^{\infty}e^{-st}{\cfrac{\mathrm{d^2}f}{\mathrm{d}t^2}}\mathrm{d}t=\color{red}{-\cfrac{\mathrm{d}}{\mathrm{d}s}\left[s^2\widetilde f(s)-sf(0)-f^{\prime}(0)\right]}=\color{blue}{-s^2\cfrac{\mathrm{d}\widetilde f}{\mathrm{d}s}-2s\widetilde f +f(0)}$$
Could someone please explain/show the steps that were taken to get from the $\color{red}{\mathrm{red}}$ expression to the $\color{blue}{\mathrm{blue}}$ expression?
Thanks.
$s^2 \tilde{f}(s)$ is the product of the functions $s^2$ and $\tilde{f}(s)$. By the product rule, the derivative of $s^2 \tilde{f}(s)$ with respect to $s$ is $$s^2 \frac{d\tilde{f}}{ds} + 2s \tilde{f}(s)$$.
$f(0)$ is constant (does not depend on $s$). The derivative of $sf(0)$ is $f(0)$. $f'(0)$ is also constant. Its derivative is zero.