Let $f(s) = \mathcal{L} \{ x(t) \}(s).$ Then
$$\mathcal{L}\{tx^{(3)}(t)\}(s)=-1\frac{d}{dx}\mathcal{L}\{x^{(3)}(t)\}(s)=-\frac{d}{ds}[s^2f(s)]=-[2sf(s)+s^2f'(s)]$$
using the identities
$$ f^{\prime}(t) \quad s F(s)-f(0) $$
and $$ t^n f(t), \quad n=1,2,3, \ldots \quad(-1)^n F^{(n)}(s). $$
Should the minus sign actually be there?
In https://math.stackexchange.com/questions/4886992/what-is-the-inverse-laplace-transform-of-sfs, it was computed that
$$\mathcal{L}^{-1}\left\{s F^{\prime \prime}(s)\right\}=t^2 f^{\prime}(t)+2 t f(t)$$
which does not have a negative sign. Is this correct?
Using differentiation under the integral and integration by parts three times, we have
$$\begin{align} \int_0^\infty t x'''(t) e^{-st}\,dt&=-\frac{d}{ds}\int_0^\infty x'''(t) e^{-st}\,dt\\\\ &=-\frac{d}{ds}\left(s^3X(s)-\sum_{n=1}^3 s^{3-n}x^{(n-1)}(0)\right)\\\\ &=-3s^2X(s)-s^3X'(s)+2sx''(0)+x'(0) \end{align}$$
If $x(0)=x'(0)=0$ then we have
$$\int_0^\infty t x'''(t) e^{-st}\,dt=-3s^2X(s)-s^3X'(s)$$