I know that they probably treated $\displaystyle f(s,t) = e^{-st} f(t)$ so the integration/differentiation thing doesn't matter, but what confuses me is when they got rid of the derivative and how the "$t$" pop out? It's taking the derivative with respect to $s$ not $t$.
Proof: Consider the identity $$\frac{dF(s)}{ds} = \frac{d}{ds} \int_0^{\infty} e^{-st} f(t) dt.$$ Because of the assumptions on $f(t)$, we can apply a theorem from advanced calculus (sometimes called Leibniz's rule) to interchange the order of integration and differentiation: $$ \begin{align} \frac{dF(s)}{ds} & = \frac{d}{ds} \int_0^{\infty} e^{-st} f(t) dt\\ & = \int_0^{\infty} \frac{d \left(e^{-st} \right)}{ds} f(t) dt\\ & = - \frac{d}{ds} \int_0^{\infty} t e^{-st} f(t) dt\\ & = - \mathcal{L} \{ tf(t) \}(s). \end{align} $$ Thus, $$\mathcal{L} \{ tf(t) \}(s) = (-1) \frac{dF(s)}{ds}$$
In general, $\frac{d}{dx} e^{ax} = a e^{ax}$
Thus, since $t$ is independent of $s$, $\frac{d}{ds} e^{-st} = -te^{-st}$
$f(t)$ doesn't depend on $s$, so it's a constant with respect to differentiation by $s$, and they pulled the negative outside the integral.
Does that make sense?