Let $p \in M$ be a smooth manifold and let $f \in C^{\infty}(M)$, and let it further be known that $df|_p$ is surjective for all $p \in M$ such that $f(p) = 0$. Now, am I correct in identifying the tangent-map $df$ with the exterior derivative $d: \Omega^{n}(M) \to \Omega^{n+1}(M)$ applied to $f$, and can I use this to say that since $d$ is $\mathbb{R}$-linear, then it follows that if $df|_p$ is a submersion, then $dg|_p$ is a submersion, where $g = f-c$, since by $\mathbb{R}$-linearity of $d$, we have that $$dg = d(f-c) = df-dc = df.$$ I think one can see that $dc = 0$ most simply by viewing $dc$ as a smooth constant function on $M$ and then using local coordinates to see that since it is constant, it will just vanish.
This question is asked in the context of the regular value theorem (I think it is called). So one can then use this to say that $f$ is a locally defining function on a neighborhood $V$ around each point $p \in M$ such that $f(p)-c = 0 \iff f(p) = c$. I.e. we have $f|_V:V \to \mathbb{R}$ (our submersion) and the set $N_c = \{p \in M| f(p) = c\}$ is a submanifold of codimension $1$, where (by another theorem) for $p \in N_c$, $$T_pN_c = \operatorname{ker}dg|_p = \operatorname{ker}df|_p = \operatorname{ker} \langle \operatorname{grad}f(p),- \rangle = (\operatorname{grad}f(p))^{\perp}.$$
So my most important question is basically if I can justify $dg = df$ because of the $\mathbb{R}$-linearity of the exterior derivative?
Thank you in advance.