= Solution
Choose the <shock wave> normal to point from $S<0$ to $S>0$:
$$
\mathbf n=\frac{\nabla S}{|\nabla S|},\qquad v_n=\mathbf v\cdot\mathbf n=-\frac{S_t}{|\nabla S|}.
$$
The latter identity follows by differentiating $S(\mathbf x_s(t),t)=0$ along the moving surface. Write $[a]=a^+-a^-$ and $[\mathbf b]=\mathbf b^+-\mathbf b^-$. The <distributional derivative of the Heaviside step function> is $\partial_tH(S)=S_t\delta(S)$ and $\nabla H(S)=\nabla S\,\delta(S)$. The regular terms cancel using the conservation equations on each side, leaving
$$
\partial_ta+\nabla\cdot\mathbf b=\left([a]S_t+[\mathbf b]\cdot\nabla S\right)\delta(S)=\left([\mathbf b]-\mathbf v[a]\right)\cdot\mathbf n\,|\nabla S|\delta(S).
$$
Thus the explicitly requested vector is
$$
\boxed{\mathbf f=[\mathbf b]-\mathbf v[a].}
$$
Only its normal component matters; adding tangential velocity to the surface parametrization changes neither result. The invariant <surface delta distribution> is $\delta_s=|\nabla S|\delta(S)$, so the formula is the <moving-interface conservation jump identity> $\partial_ta+\nabla\cdot\mathbf b=([\mathbf b]\cdot\mathbf n-v_n[a])\delta_s$.
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