Frattini subgroup is normal-monotone

partial answer: Thanks to a comment I realise my answer is only sufficient in the case $\Phi(N)$ is finitely generated.

You don't need every proper subgroup of $N$ to be contained in a maximal subgroup of $N$ to reach $N=\Phi(N)(N\cap M)$.

If $M$ is a maximal subgroup of $G$ not containing $\Phi(N)$ then $G=\Phi(N)M$ so by the modular law for groups we have $$N=N\cap\Phi(N)M=\Phi(N)(M\cap N)$$

Edit:

This is a contradiction when $\Phi(N)$ is finitely generated because the Frattini subgroup of $N$ is the set of non-generators of $N$. That is $N=\langle \Phi(N),M\cap N\rangle$ implies that $N=M\cap N$ so $N\le M$. Hence $G=\phi(N)M\le NM=M$.