$\left[ - \mathcal{L} \left\{ \dfrac{F(t)}{t} \right\} \right]_s^\infty$ to $\mathcal{L} \left\{ \dfrac{F(t)}{t} \right\}\bigg\vert_s$

This seems like mostly a confusion about notation. The notation

$$\left[f(x)\right]_s^\infty$$

is shorthand for

$$\lim_{b\to\infty} f(b)-f(s)\ .$$

So

$$\left[- \mathcal{L} \left\{ \dfrac{F(t)}{t}\right\}\right]_s^{\infty} = -\lim_{b\to\infty}\mathcal{L}\left\{\frac{F(t)}{t}\right\}\bigg\vert_b + \mathcal{L}\left\{\frac{F(t)}{t}\right\}\bigg\vert_s$$

Because of the assumption that the Laplace transform approaches $0$ as the input approaches $\infty$ the first term vanishes. The second is just the value of the Laplace transform of $\frac{F(t)}{t}$ evaluated at $s$. Both

$$\mathcal{L}\left.\left\{\frac{F(t)}{t}\right\}\right\vert_s$$

and

$$\mathcal{L}\left\{\frac{F(t)}{t}\right\}$$

are ways to denote this. In the second case the dependence on $s$ has been suppressed.