Assuming $M$ is a manifold of dimension $\dim M = n+k$ and $F \colon M \to \mathbb{R}^n$ is a smooth function, such that the levelsets $F^{-1}({c})$ are submanifolds of dimension $k$.
Can I therefore conclude, that $F$ has full rank on an open and dense subset of $M$?
Using the Edit, my question seems more like: Under which assumptions do I find "enough" clean values, such that $F$ has full rank on an open and dense subset of $M$?
Edit: I think in this generality it isn't true. I have to assume, that $F(x) =c \in \mathbb{R}^n$ is a clean value, i.e. $\ker d_xF = T_c \left(F^{-1}(c)\right)$ for all $x \in F^{-1}(c)$.
Now if $\gamma \colon I \to M$ is a path contained in $F^{-1}(c)$, then $\dot \gamma (0) \in \ker d_xF$. So $T_c \left(F^{-1}(c)\right) \subset \ker d_xF$ and since $\dim F^{-1}(c) = k$, we know $\dim \ker d_xF \geq k$ and so $\dim d_xF(T_xM) \leq n$.
If $c \in \mathbb{R}^n$ is a clean value, then $\dim \ker d_xF =k$ and so $F$ is a submersion for all $x \in F^{-1}(c)$.
Otherwise let $M = \mathbb{R}^2$ and $F \colon \mathbb{R}^2 \to \mathbb{R}, \ (x,y) \mapsto x^2$. Then $F^{-1}(c) = \{(c,y) | y \in \mathbb{R} \}$ and so $\dim F^{-1}(c) = 1$ for all $c$. But in $(0,0)$ we have that $\ker d_xF = \mathbb{R}^2$, so $F$ doesn't have full rank for $(0,0)$.