Normal Hall subgroups of $M$-groups are $M$-groups. Dornhoff proved this theorem(Theorem 4.1. on page 250) in the paper https://link.springer.com/article/10.1007/BF01109806.
The first few lines of the proof go as follows:

On the contrary, assume $G\triangleleft M$, where $M$ is an $M$-group and $G$ is a normal Hall subgroup of $M$, and $\theta$, an irreducible character of $G$, is not monomial. Clifford's Theorem tells us that $(\theta^M)_G=k(\theta_1+\cdots+\theta_l)$, $\theta_1=\theta$, the $\theta_i$ distinct irreducible characters of $G$, $kl=|M:G|$, $k=((\theta^M)_G,\theta)=(\theta^M,\theta^M)$.

Question:
I do not understand why we can apply the Clifford's theorem in this way, namely, the Clifford's theorem deals with the restriction of an irreducible character to some normal subgroup. But $\theta^M$ is not guaranteed to be irreducible in $M$ above. Then how can we still apply the Clifford's theorem in this case? Also, why is $kl=|M:G|$? $l$ should equal the index of the inertia subgroup of $\theta$ in $M$, so equivalently, why does $k$ equal the index of $G$ in the inertia subgroup of $\theta$ in $M$?

Edit: I am trying to prove this theorem with the ideas from the original proof but using Clifford's theorem in a different way with the author.
Suppose $N$ is a normal Hall subgroup of the $M$-group $G$. Let $\theta\in Irr(N)$, we want to show that $\theta$ is monomial.
Let $T=I_G(\theta)\supseteq N$ be the inertia subgroup of $\theta$ in $G$, then $\theta$ is invariant in $T$. Moreover, $N$ is a normal Hall subgroup of $T$ since $\gcd(|N|,|G/N|)=1$ and $|T/N|\mid |G/N|$. By Theorem 6 of Gallagher, $\theta$ extends to an irreducible character of $T$, i.e., $\theta=\chi|_N$ for some $\chi\in Irr(T)$.
By theorem 6.11 of Isaacs's character theory of representation theory, we know that $\chi^G\in Irr(G)$ and $$ 1=(\theta,\theta)_N=(\chi|_N,\theta)_N=((\chi^G)|_N,\theta)_N $$
Note that
$$((\chi^G)|_N,\theta_i)_N=((\chi^G)|_N,\theta)_N=1, \quad i=1,...,|G:T|\text{ and }\theta_1=\theta$$
where $\theta_i$'s are all the distinct conjugate of $\theta$. By Clifford's theorem, we know that $$(\chi^G)|_N=\sum_{i=1}^{|G:T|}\theta_i.$$
Since $\chi^G\in Irr(G)$, it is monomial, say $\chi^G=\lambda^G$ where $\lambda\in Irr(H)$ for some $H\le G$ and $\lambda(1)=1$. Consider
$$ (\lambda^{NH})|_N=(\lambda|_{N\cap H})^N. $$
(Not finished)

By the Schur–Zassenhaus theorem, we can write $G=NH$. If we can always take $H$ to be the complement of $N$, I think we will be done.
 A: You need the following lemmata.
(I urge you to draw a diagram to depict the different groups and characters.)
Lemma 1 Let $H$ and $K$ be subgroups of $G$ and $\varphi$ a character of $H$. Assume that $(\varphi^G)_K \in Irr(K)$. Then $G=HK$.
Proof Put $\varphi^G=\chi$. Observe that since $(\varphi^G)_K$ is irreducible, both $\varphi$ and $\chi$ must be irreducible. Let $\psi \in Irr(H \cap K)$ be an irreducible constituent of $\varphi_{H \cap K}$. By Frobenius Reciprocity, $\varphi$ must be an irreducible constituent of $\psi^H$, say $\psi^H=a\varphi+\Delta$, with $a$ a positive integer and $\Delta$ a character of $H$ with $[\Delta,\varphi]=0$  or $\Delta=0$. It follows that $\psi^G=(\psi^H)^G=a\varphi^G+\Delta^G=a\chi+\Delta^G$. So again by Frobenius Reciprocity, $[\psi^G,\chi]=[(\psi^K)^G,\chi]=[\psi^K,\chi_K] \geq a.$ 
On the other hand, $a=[\varphi,\psi^H]=[\varphi_{H \cap K},\psi]$, so $\varphi_{H \cap K}=a\psi+\Gamma$ with $\Gamma$ a character of $H \cap K$ with $[\Gamma,\psi]=0$  or $\Gamma=0$. Hence $(\varphi_{H \cap K})^K=a\psi^K+\Gamma^K$ and it follows that $[(\varphi_{H \cap K})^K, \chi_K]=[a\psi^K+\Gamma^K,\chi_K]=a[\psi^K,\chi_K]+[\Gamma^K,\chi_K] \geq a^2$, by the last formula of the previous paragraph. This means that the irreducible character $\chi_K$ has at least multiplicity $a^2$ as an irreducible constituent of $(\varphi_{H \cap K})^K$. This implies that
$$(\varphi_{H \cap K})^K(1)=\varphi(1)|K:H \cap K| \geq a^2\chi(1) \geq \chi(1)=\varphi(1)|G:H|$$ It follows that $|G| \leq \frac{|H| \cdot |K|}{|H \cap K|}=|HK|$. Hence, $G=HK$, since $HK \subseteq G$ as a set.$\square$
Notes (1) This is Problem(5.7) in Isaacs' book. (2) From the last part of the proof it follows that in fact $a=1$.
Corollary 1 Let $\chi$ be a monomial character of $G$ and suppose that $K \leq G$ with $\chi_K \in Irr(K)$. Then $\chi_K$ is monomial.
Proof We can find an $H \leq G$ and a linear $\lambda \in Irr(H)$ with $\lambda^G=\chi$. Since $\chi_K$ is irreducible, Lemma 1 guarantees that $G=HK$. But then (see Problem(5.2) in Isaacs's CTFG), $\chi_K=(\lambda^G)_K=(\lambda_{H \cap K})^K$, whence $\chi_K$ is monomial.
Lemma 2 Let $N \unlhd G$ with $G/N$ solvable. Let $\vartheta \in Irr(N)$ and $\chi \in Irr(G|\vartheta)$. Then $\frac{\chi(1)}{\vartheta(1)} \mid |G:N|$.
Proof Use induction on $|G:N|$. We can assume that $G/N$ is non-trivial and hence there exists a subgroup $M/N \lhd G/N$ with $|G/N:M/N|=|G:M|=p$, with $p$ a prime. We can choose a $\varphi \in Irr(M)$ lying under $\chi$ and lying above $\vartheta$. By Clifford's Theorem (hence Corollary (6.19) in CTFG) we have either $\chi_M=\varphi$ or $\chi_M=\sum_{i=1}^p \varphi_i$. Hence either $\chi(1)=\varphi(1)$ or $\chi(1)=p\varphi(1)$. By induction in the first case we get $\frac{\varphi(1)}{\vartheta(1)}=\frac{\chi(1)}{\vartheta(1)}  \mid |M:N| \mid |G:N|$. In the latter case $\frac{\chi(1)}{\vartheta(1)}=\frac{\chi(1)}{\varphi(1)} \cdot \frac{\varphi(1)}{\vartheta(1)}=p\frac{\varphi(1)}{\vartheta(1)} \mid p|M:N|=|G:N|$ and the proof is complete.
Note This is Exercise(6.7) in Isaacs' book CTFG.
Corollary 2 Let $N \unlhd G$ with $G/N$ solvable. Let $\chi \in Irr(G)$ with gcd$(\chi(1),|G:N|)=1$, then $\chi_N \in Irr(N)$.
Proof If $\vartheta \in Irr(N)$ lies under $\chi$, then $\frac{\chi(1)}{\vartheta(1)}$ divides both $\chi(1)$ and $|G:N|$. So $\chi(1)=\vartheta(1)$ and by Clifford's Theorem $\chi_N=\vartheta$.
Theorem (L. Dornhoff) Let $G$ be an $M$-group and $N$ a normal Hall-subgroup. Then $N$ is again an $M$-group.
Proof Let $\vartheta \in Irr(N)$ and $\chi \in Irr(G|\vartheta)$. Then there exists an $H \leq G$ and a linear $\lambda \in Irr(H)$ with $\chi=\lambda^G$. Note that $NH$ is a subgroup ($N$ is normal), also $(\lambda^{NH})^G=\lambda^G=\chi$, implying  $\lambda^{NH} \in Irr(NH)$. Now, $\lambda^{HN}(1)=|NH:H|=|N:N \cap H| \mid |N|$. Note $|NH:N| \mid |G:N|$ is relatively prime to $|N|$, so, since $G$ is solvable (Taketa's Theorem), Corollary 2 yields $(\lambda^{NH})_N \in Irr(N)$. Since $(\lambda^{NH})_N=(\lambda_{N \cap H})^N$, it follows that $(\lambda^{NH})_N$ is monomial. Observe that $(\lambda^{NH})_N$ lies under $\lambda^{NH}$ and in its turn, $\lambda^{NH}$ lies under $\chi$. But $\vartheta \in Irr(N)$ and $\vartheta$ lies under $\chi$, so by Clifford's Theorem, $(\lambda^{NH})_N$ and $\vartheta$ are conjugate in $G$. The monomiality of $(\lambda^{NH})_N$ now implies that of $\vartheta$.
