It's using the fact that
[tex]\mathfrak{L}\left( \sum a_i \frac{d^k}{dx^k} y \right) = \sum a_i \mathfrak{L} \left( \frac{d^k}{dx^k} y \right)[/tex]
And that if you take the Laplace operator of the kth derivative of y you get sk L(y) plus some values of y and its derivatives at 0 (more specifically the general differentiation property at http://en.wikipedia.org/wiki/Laplace_transform#Properties_and_theorems)
The ak coefficients that were next to the differential operators stick around and multiply the sk L(y) guys, meaning you get exactly q(s) out if you started with q(D), but the polynomial terms depend on the derivatives at zero so is hard to calculate what its relationship with q(s) is