Let $F:N\rightarrow M$ be a smooth embedding. Then, $F(N)$ is an embedded submanifold of $M$.
So what I have tried is the following:
Let $F(p)\in F(N)$ for some $p\in N$. As $F$ is an immersion, by the immersion theorem, there exists charts $(U,\phi)=(U,x^1,….,x^n)$ about $p$ and $(V,\psi)=(V,y^1,…..,y^m)$ about $F(p)$ on which $F(U)\subseteq V$ and $(\psi \circ F\circ \phi^{-1})(x^1,…..,x^n)=(x^1,…..,x^n,0,…..,0)$ on $\phi(U)$.
As $F(N)$ is homeomorphic to $N$, there exists an open set $W$ in $M$ such that $F(U)=V'\cap F(N)$. Hence, $V\cap V'\cap F(N)= V\cap F(U)=F(U)$.
Claim: $\psi(V\cap V'\cap F(N))$ = $\psi(V\cap V')\cap (\mathbb{R}^{n}$ $ \times \{0\} )$
Now clearly, $\subseteq$ holds. However, I don't think the reverse inclusion holds. I mean, for $\psi(q)\in \psi(V\cap V') \cap (\mathbb{R}^n \times \{0\})$ , there is no immediate reason for $q$ to be in $F(N)$ as well.
To remedy this situation, I suppose we would need to construct a new chart $W$ such that $W\subseteq V\cap V'$ and $W\subseteq F(N)$. I would like a nudge in the right direction.
Best Answer
The inclusion $\psi(V\cap V'\cap F(N))\subseteq\psi(V\cap V')\cap (\mathbb{R}^{n}\times \{0\} )$ could indeed be strict, but note that if dim$(N)$<dim$(M)$ restricting the chart $(V,\psi)$ to some open set $W$ with $W\subseteq F(N)$ cannot work for dimensional reasons.
To get rid of the possible strict inclusion just further rectrict to
$$W=V\cap V'\cap \psi^{-1}(\phi(U)\times\mathbb R^{m-n})$$
This ensures that in addition the ''$\mathbb R^n$-part" in the image of $\psi$ only comes from points in the domain of $\phi$. More formally, as an injection $\psi$ respects intersections, so $$\psi(W)\cap(\mathbb R^n\times\{0\}) =\psi(V\cap V')\cap(\phi(U)\times\mathbb R^{m-n})\cap (\mathbb R^n\times\{0\}) =\psi(V\cap V')\cap(\phi(U)\times\{0\}) =\psi(V\cap V')\cap\psi(F(U)) =\psi(V\cap V'\cap F(N))$$ and intersecting both sides with $\psi(W)$ yields
$$\psi(W)\cap(\mathbb R^n\times\{0\})=\psi(W\cap F(N))$$
which finishes your proof.