First of all, let us make a few general considerations about the existence of such an $a(y)$. That amounts to asking whether the Radon-Nikódym derivative of $H^s\circ D$ with respect to $H^s$ exists. If $H^s$ and $H^s\circ D$ were $\sigma$-finite, by the Radon-Nikódym theorem this is the same as asking whether $H^s\circ D$ is absolutely continuous with respect to $H^s$: if $H^s(A)=0$, then $H^s(DA)=0$ for any $H^s$-measurable $A$. The problem is that if $s<n$, then $H^s$ is not $\sigma$-finite on $\mathbb{R}^n$, so we cannot appeal to the Radon-Nikódym theorem for existence unless $s=n$.
Now, if $D$ is an homothety, i.e. $D$ is of the form $D=\lambda U$ where $\lambda>0$ and $U$ is an isometry (i.e. an orthogonal matrix), then $H^s(DA)=\lambda^sH^s(A)$, hence in this case $a(y)=\lambda^{-s}$. If, however, $D$ is not an homothety - e.g. a symmetric, non-singular but non-orthogonal matrix with $\det D=1$ -, I do not know of any simple transformation formula for the $s$-dimensional Hausdorff measure, and I suspect that there may be none leading to a formula such as the one you are looking for.
The problem is that since such a linear map might expand certain directions while contracting others, the naive expectation that $H^s(DA)=|\det D|^{\frac{s}{n}}H^s(A)$ (where $n$ is the rank of $D$) coming from the case that $D$ is an homothety may fail short depending on how $A$ is oriented with respect to the principal axes of $D$, as shown here. Since one can change this relative orientation just by an isometry (which, as seen above, does not change Hausdorff measure), I do not see any chance of such a function $a(y)$ existing independently of the choice of $A$. More precisely, if we take $f\equiv 1$, then your last formula just becomes $$D_*H^s(DA)=H^s(A)=\int_{DA}a(y)\;\mathrm{d}H^s(y)$$ for all $A\subset\mathbb{R}^n$ $H^s$-measurable. If we replace $A$ with $UA$ where $U$ is a suitable isometry, we should have $$H^s(UA)=H^s(A)=\int_{D(UA)}a(y)\;\mathrm{d}H^s(y)\ ,$$ so $a(y)$ should have an "anisotropic" behavior which seems incompatible with the fact that it is a scalar function. I do not know if I was able to make my concerns sufficiently precise - if I can find a more explicit example, I will update my answer.
