Every finite measure on a topological space is outer regular if and only if it is inner regular, so compactness would be sufficient (and every finite measure on metric space is automatically both inner and outer regular). Another sufficient condition is that the space is the union of an increasing sequence of compact sets, which covers the case of $\mathbb{R}^n$.
One can look this up in: Aliprantis & Border 2006, "Infinite Dimensional Analysis" in section 12.1.
Edit: Lurker has pointed out that aliprantis and Border use the weaker notion of inner regularity of apprximating sets from below by closed sets. What is called inner regularity here, they call tightness. It is automatic for finite measures on Polish spaces but not all metric spaces.
Disintegration of the volume w.r.t. a surface measure
on $\Sigma_s=\{F=s\}\cap D$ and the parameter $s$ is difficult when
$\Sigma_0=\partial D$ is only $C^1$. $\Sigma_s$ is in general only Lipschitz,
although with a Lipschitz constant that goes to 1 as $s\rightarrow 0$.
I guess there is still a well-defined surface measure on $\Sigma_s$
but I wouldn't know how to use it in calculations.
It is much easier to work directly with the volume form:
In the following, write $n_p$ for the unit vector, normal to the tangent space of $\Sigma_0$ at $p$ and
pointing inwards in $D$.
First a geometric estimate:
For $\delta>0$ consider the quantity:
$$ \alpha(\delta) = \{ n_p \cdot v :
p\in \Sigma_0, v \in T_q \Sigma_0, |v|=1, d(p,q)\leq \delta\}.$$
The dot-product is the cosine of the angle between the normal vector $n_p$ at $p$
and a tangent vector at $q$ when $p,q\in \Sigma_0$ are $\delta$-close. $\alpha$ estimates
the local 'flatness' of the surface.
We have $\alpha(0)=0$ and $\lim_{\delta\rightarrow 0} \alpha(\delta)=0$
because $\Sigma_0$ is assumed $C^1$ and compact.
Let $f$ be a continuous function with
support in a small ball $B(p,\delta/2)$ around
$p\in \Sigma_0$. We will
show that for $\delta$ small
$$ \lim_{s\rightarrow 0 } \frac{1}{s} \int_{\{0\leq F \leq s\}\cap D} f\; d^d x =
\int_{\Sigma_0} f\; d\sigma_0$$
where $d\sigma_0$ is the area measure on the surface (well-defined as $\Sigma_0$ is $C^1$).
This suffices since for general $f$
you may use a partition of unity to reduce to this case.
Local coordinates:
Choosing $\delta>0$ small enough we may ensure that $B(p,\delta)\cap \Sigma_0$
is the (embedded) image of a $C^1$ map:
$$ j : U \rightarrow \Sigma_0 \subset {\Bbb R}^d, \; 0\in U \subset {\Bbb R}^{d-1}, j(0)=p, $$
where $j'(u)$ is injective for all $u\in U$ and its image is the tangent space
of $\Sigma_0$ at $j(u)$. Write $n(u) = n_{j(u)}$ and
consider now the map:
$$ \Phi(u,s) = j(u) + s n(u) , u\in U, s\in [0,\delta/2]$$
The Jacobian determinant, $J(u,s) = \det \Phi' = \det ( j'(u) + s n'(u), n(u))$
is non-zero at $(u,s)=(0,0)$ and since continuous in
$u$ and $s$ we may by the inverse function theorem (possibly choosing $\delta$
smaller) assume that $\Phi$ is a bijection onto its image.
We then have an expression
for the volume element $d^d x = J(u,s) d^{d-1}u ds$ in local coordinates.
Furthermore, when restricted to $s=0$ it gives the $(d-1)$-dim area element on the surface
(because $n$ is a unit normal to the tangent space):
$$ d\sigma_0 = J(u,0) d^{d-1} u $$
The distance from $\Phi(u,s)$ to $\Sigma_0$ is clearly at most $s$ but a
geometric consideration (make a drawing) using the definition of $\alpha$ shows that it is
not smaller than $s \sqrt{1-\alpha(\delta)^2}$ (the last factor is an upper
bound for the sine of the angle between two normal vectors in $\Sigma_0\cap B(p,\delta)$).
It follows that for $0\leq s \leq \delta/2$ and restricting
to the set $B(p,\delta/2)$ we have
$$ \{x\in D: 0\leq F\leq s \sqrt{1-\alpha(\delta)^2}\} \subset \Phi(U,s)
\subset \{x\in D: 0\leq F\leq s\} . $$
Therefore,
$$
\frac{1}{s} \int_{0\leq F\leq s \sqrt{1-\alpha(\delta)^2} } f \; d^d x \leq
\frac{1}{s} \int_0^s \left( \int_U f(u,s) J(u,s) d^{d-1}u \right) ds \leq
\frac{1}{s}\int_{0\leq F\leq s} f \; d^d x
$$
As $s\rightarrow 0^+$ the middle term goes to $\int_{\Sigma_0} f\; d\sigma_0$ by simple continuity,
and the outer terms (limsup and liminf)
agree within a factor $ \sqrt{1-\alpha(\delta)^2}$.
Patching things together we may
finally let $\delta$ go to zero and conclude since this
factor goes to one.
Best Answer
You are right. They invoke $\mathcal{H}^s =\mathcal{L}^s$ on $\mathbb{R}^s$ for $s \in \mathbb{N}$. (And then the fact that Lebesgue measure splits as product of lower dimensional Lebesgue measures.)
Use of $\mathcal{H}^s$ is motivated (actually necessitated) by the fact that in the general coarea formula your preimages will not be identifiable with Euclidean spaces in any canonical way, so this notation is adopted in prediction of that and so that at the end one can compare and contrast.
Finally, you need Lebesgue measure in applying Fubini's theorem and also we like to integrate things against Lebesgue measure rather than Hausdorff measure, although ultimately they coincide.