I normally don't like to cite my own work on MO, but this time the preprint arXiv:1803.04064 was written, together with L. Caputo, having the OP's question in mind; and so, first of all, let me thank Alex for having asked it.
The main result is a purely algebro-arithmetic proof that for any base field $k$ and for every dihedral extension $F/k$ of degree $2q$ with $q$ odd, it holds
$$
\frac{h(k)^2h(F)}{h(L)^2h(K)}=\frac{\lvert\widehat{H}^0(D_{2q},\mathcal{O}_F^\times\otimes\mathbb{Z}[1/2])\rvert}{\lvert\widehat{H}^{-1}(D_{2q},\mathcal{O}_F^\times\otimes\mathbb{Z}[1/2])\rvert}
$$
in the OP's notations (which are different from the ones used in the preprint). When I say that the proof is ``algebro-arithmetic'' I mean that the main ingredient is class field theory and some group cohomology: no $\zeta$- or $L$-functions are involved, neither is the classification of integral representation of dihedral groups. The key point is that, if we call $G_q=\operatorname{Gal}(F/K)\cong\mathbb{Z}/q\subseteq D_{2q}$, then $G_q$-cohomology of a $D_{2q}$-module (typically: units, ad`eles, local units, etc.) has an action of $\operatorname{Gal}(K/k)$ and when the module is uniquely $2$-divisible, this induces identifications $\widehat{H}^i(G_q,-)^+=\widehat{H}^i(D_{2q},-)$ as well as $\widehat{H}^i(G_q,-)^-=\widehat{H}^{i+2}(D_{2q},-)$, where $\pm$ are the eigenspaces with respect to the action of $\operatorname{Gal}(K/k)$. Therefore we can use class field theory for the abelian extension $F/K$ to deduce information about cohomology groups for $D_{2q}$.
As corollary of the above formula we show that, for every prime $\ell$, the following bounds hold
$$
-av_\ell(q)\leq v_\ell\left(\frac{h(k)^2h(F)}{h(L)^2h(K)}\right)\leq bv_\ell(q)
$$
where $a=\operatorname{rank}_\mathbb{Z}\mathcal{O}_K^\times + \beta_K(q) + 1$, $b=\operatorname{rank}_\mathbb{Z}\mathcal{O}^\times_{k}+\beta_k(q)$ and $v_\ell$ denotes the $\ell$-adic valuation; the "defect" $\beta_M(q)\in\{0,1\}$ (for $M=K,k$) is defined to be $1$ if $\mu_M(q)$ is non-trivial, and $0$ otherwise. From this, we deduce even sharper bounds in case $K$ is either CM (with totally real subfield equal to $k$) or if it is totally real. Since when $k=\mathbb{Q}$ this is always the case, we prove as a special result that in every dihedral extension of $\mathbb{Q}$ of degree $2p$ the formula required by the OP holds, again without resorting to any analytic or ``hard'' representation-theoretic result. Actually, restricting to the prime case $q=p$ has no utility whatsoever, and when $F/\mathbb{Q}$ is any dihedral extension of degree $2q$ ($q$ odd!) we deduce that the ratio of class numbers verifies
$$
0 \geq v_\ell\left(\frac{h(F)}{h(K)h(L)^2}\right)\geq
\begin{cases}
-2&\text{if $K$ is real quadratic}\\
-1&\text{if $K$ is imaginary quadratic}
\end{cases}
$$
where, again, $v_\ell$ is the $\ell$-adic valuation. It is perhaps interesting to observe that ramification plays little to no role in our proof (ramification indexes only appear as well-controlled orders of cohomology groups of local units) and so assuming particular ramification behaviours wouldn't simplify it.
The $p$ dimensional simples are the inductions of the 1-d reps of $C_q$. That is, the generators of order $p$ and $q$ act by the matrices
$A=\begin{bmatrix} 0 & 0 & \cdots &0
& 1\\
1 & 0 & \cdots &0& 0\\
0& 1 & \cdots &0& 0\\
\vdots&\vdots &\ddots&\vdots &\vdots\\
0 & 0& \cdots &1& 0
\end{bmatrix}\,$ and $\,B=\begin{bmatrix}
\zeta & 0 & \cdots &0& 0\\
0& \zeta^b & \cdots &0& 0\\
\vdots&\vdots &\ddots&\vdots &\vdots\\
0 & 0& \cdots &\zeta^{b^{p-2}}& 0\\
0 & 0 & \cdots &0
& \zeta^{b^{p-1}}\\
\end{bmatrix}\,$
for $\zeta$ any $q$th root of unity, and $b$ an element with multiplicative order $p$ in $\mathbb{Z}/q\mathbb{Z}$. There are $d$ distinct ones, since these will be isomorphic if the same root of unity appears in the diagonal of $B$ in both.
Best Answer
It depends what that person meant by "Mackey Machine". Mackey decomposition, which tells you how to decompose a module induced from one subgroup when restricted to another subgroup works over pretty much any ring. Clifford's theorem, which tells how an irreducible representation of a group $G$ restricts to a normal subgroup $N$ (as a fixed multiple of a sum of $G$-conjugate irreducible representations of $N$) does not really need algebraic closure of the ground ring. If the $G$-conjugate irreducible representations of $N$ form an orbit of size greater than one, then one can reduce the situation to an intermediate subgroup, strictly containing $N$, but smaller than $G$. However, when we have an irreducible $G$-module whose restriction to $N \lhd G$ is a multiple of a single irreducible module $U$ for $N,$ then the analysis can become delicate when working over a field $F$ which is not algebraically closed. The reason is that the $N$-endomorphism ring of $U$ might not just consist of scalars, as would be the case if we were working over an algebraically closed field. We do have an action of $G$ on ${\rm End}_{F}(U),$ which is induced by inner automorphims of the matrix ring. However, this asscoiates an inner automorphism to $g \in G$ which is only determined up to multiiplication by an element of ${\rm End}_{FN}(U).$ Hence we do not necessarily get a homomorphism from $G/N$ to ${\rm PG}L(m,F)$ where $m = {\rm dim}_{F}(U)$ from the original representation.
Later edit: I don't know if the question has changed, but I just read it again. The statement about he group algebra doesn't mention $s$, only $r$, but then asks what $s$ would be. That may just be a typo. As long as $p^{\prime}$ is odd and different from $p,$ the group algebra $FG$ is still semisimple for $G$ dihedral with $2p^{\prime}$ elements and $F = \mathbb{Z}/p\mathbb{Z}.$ The number of isomorphism types of simple $FG$-module can be calculated in that case. Let $C$ be the cyclic subgroup of index $2$ of $G.$ A faithful simple $FG$-module $V$ can restrict in several ways to and $FC$-module. It can remain simple or it can restrict as two copies of a single simple. If $C = \langle c \rangle,$ and $c^{-1}$ can not be written in the form $c^{p^{t}}$ for some integer $t,$ then the split will be as two different simples; otherwise it will remain simple, or it might split as a direct sum of two copies of a single simple for $FN.$ But the latter cas can't occur, otherwise the endomorphism ring ${\rm End}_{FG}(V)$ is a finite division ring, using the more precise version of Schur's Lemma, henceis a field. On the other hand ${\rm End}_{FN}(V)$ is isomorphic to a $2 \times 2$-matrix ring over a field, and is acted on by $G/C$, a group of order $2,$ as an inner automorphism. The inner automorphism can't be scalar, as its fixed points are a field, wheeas they are the whole matrix ring, a contradiction. Nor can it be non-scalar of order $2$, otherwise its fixed-points contain non-trivial idempotents, again contrary to its fixed points being a field. Once the faithful simple modules are dealt with, the non-faithful one follow using (mathematical) induction. For a more general semidirect product, say $G = NT$, then it is necessary to understand how $T$ acts on various matrix ring over fields, ( if working with simple $FG$-modules for finite fields $F$), the fields in question being fields of the form ${\rm End}_{FN}(U)$ for simple $FN$-modules $U$. That's assuming characteristic of $F$ coprime to $|G|$. If the characteristic of $F$ is not coprime to $|G|$, the simple $FG$ modules can be described in similar terms, but the group algebra is of course no longer semisimple.