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For derivative controller transfer function, it's \$K_ds\$ but the Laplace Transform of \$\frac{df(x)}{dx}\$ is \$sF(s)-f(0)\$. For example if
$$f(x) = \frac{dg(x)}{dx}$$ hence its Laplace will be
$$F(s) = sG(s)-g(0)$$
If the transfer function of derivative is \$s\$ then
$$\frac{F(s)}{G(s)}=s \Leftrightarrow F(s) = sG(s)$$
how can inverse Laplace transform of \$F(s) = sG(s)\$ give \$f(x) = \frac{dg(x)}{dx}\$
The bilateral Laplace Transform of \$ f(t) \$ is \$ F(s) = sG(s) \$
See: https://en.wikipedia.org/wiki/Two-sided_Laplace_transform#Properties
It is due to the fact that there is an assumption that all initial conditions are zero. This is necessary, as the transfer function is based on linear time invariant systems. If the initial conditions are not zero, there will be a non-linear effect present, due to adding constants.
