= Solution
Let $P=L\ltimes U$ and $Q=M\ltimes V$ be rational <parabolic subgroups> with rational <Levi subgroups>. Work with complex representations of their <finite groups of Lie type>. Here <Harish-Chandra induction> and <Harish-Chandra restriction> mean
$$
R_{L\subset P}^{G}(E)=\operatorname{Ind}_{P^F}^{G^F}\operatorname{Inf}_{L^F}^{P^F}E,\qquad {}^*R_{M\subset Q}^{G}(A)=A^{V^F}.
$$
The latter is an $M^F$-module, and it is adjoint to the former by <Frobenius reciprocity>.
Define
$$
\mathcal S(M,L)=\{x\in G:M\cap{}^xL\text{ contains a maximal algebraic torus of }G\}.
$$
For $x\in\mathcal S(M,L)^F$, put $H_x=M\cap{}^xL$ and let $\operatorname{ad}x$ transport an $L^F$-module to a $({}^xL)^F$-module. The <Mackey formula for Harish-Chandra induction> is the natural functor isomorphism
$$
\boxed{{}^*R_{M\subset Q}^{G}R_{L\subset P}^{G}\cong\bigoplus_{x\in M^F\backslash\mathcal S(M,L)^F/L^F}R_{H_x\subset M\cap{}^xP}^{M}\ {}^*R_{H_x\subset Q\cap{}^xL}^{{}^xL}\ \operatorname{ad}x.}
$$
On <characters> the direct sum becomes a sum. The intersection parabolics displayed in the subscripts are important: they specify exactly which <Harish-Chandra induction> and <Harish-Chandra restriction> occur.
We use the following parabolic-intersection lemmas, stating explicitly the geometric input to the proof. First, representatives of $M^F\backslash\mathcal S(M,L)^F/L^F$ represent exactly the double cosets $Q^F\backslash G^F/P^F$. Second, for such $x$, the groups $M\cap{}^xP$ and $Q\cap{}^xL$ are <parabolic subgroups> of $M$ and ${}^xL$, respectively, with common <Levi subgroup> $H_x$ and respective <unipotent radicals> $M\cap{}^xU$ and $V\cap{}^xL$. If $K=Q\cap{}^xP$, projection $Q\to M$ maps $K$ onto $M\cap{}^xP$ with kernel $K\cap V$. Projection ${}^xP\to{}^xL$ maps $K\cap V$ onto $V\cap{}^xL$ with kernel $V\cap{}^xU$. These assertions hold on rational points as well. The surjectivity on rational points uses connectedness of the unipotent kernels and the <Lang theorem for algebraic groups>. These are the compatible-Levi and rational double-coset lemmas; they follow by applying the root-subgroup decomposition after choosing the common <maximal algebraic torus>.
For completeness, the representation-theoretic reduction is as follows. The <group algebra> realization of <induced representation> splits according to the double cosets:
$$
\mathbb C[G^F]=\bigoplus_x\mathbb C[Q^FxP^F]
$$
as a $(Q^F,P^F)$-bimodule. Tensoring on the right with an inflated $L^F$-module $E$ identifies each summand with induction from $K^F=(Q\cap{}^xP)^F$. This proves the finite-group <Mackey restriction formula> in the form
$$
\operatorname{Res}_{Q^F}^{G^F}R_{L\subset P}^{G}(E)\cong\bigoplus_x\operatorname{Ind}_{K^F}^{Q^F}\left(\operatorname{Res}_{K^F}^{({}^xP)^F}{}^x\widetilde E\right).
$$
Here ${}^x\widetilde E$ is trivial on $({}^xU)^F$.
We next record the elementary normal-subgroup identity underlying <Harish-Chandra restriction>. If $N\triangleleft D$ are finite groups, with $N$ a <normal subgroup>, and $C\subset D$, then for a complex $C$-module $A$,
$$
\left(\operatorname{Ind}_C^D A\right)^N\cong\operatorname{Ind}_{CN/N}^{D/N} A^{C\cap N}.
$$
To see this, replace invariants by coinvariants using the averaging idempotent $|N|^{-1}\sum_{n\in N}n$. Taking coinvariants in $\mathbb C[D]\otimes_{\mathbb C[C]}A$ first replaces $\mathbb C[D]$ by $\mathbb C[D/N]$, and forces $C\cap N$ to act trivially on $A$. Thus it yields $\mathbb C[D/N]\otimes_{\mathbb C[CN/N]}A_{C\cap N}$; averaging over $C\cap N$ identifies this last module with the one displayed. This also shows why characteristic zero makes the argument exact.
Apply this identity to $D=Q^F$, $N=V^F$, $C=K^F$. By the intersection lemmas, $CN/N=(M\cap{}^xP)^F$. Since ${}^x\widetilde E$ is trivial on $({}^xU)^F$, its $(K\cap V)^F$-invariants are exactly
$$
({}^xE)^{(V\cap{}^xL)^F}={}^*R_{H_x\subset Q\cap{}^xL}^{{}^xL}({}^xE).
$$
Moreover $(M\cap{}^xU)^F$ acts trivially on this module, so induction from $(M\cap{}^xP)^F$ to $M^F$ is precisely <Harish-Chandra induction> from $H_x$. Substituting into each double-coset summand gives the boxed formula. This proves the formula for rational parabolics; it does not assume a general Mackey formula for nonrational <Deligne-Lusztig induction>.
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