The tangent space of a regular level set is the kernel

Because $q$ is a regular value and $p\in F^{-1}(q)$, [F1] makes $F$ a submersion at $p$. Write $m:=\dim M$, $n:=\dim N$, and $\mathrm{\ell }:=m-n$. By [L2], choose local coordinates near $p$ and $q$ in which the representative of $F$ is $(u,v)\mapsto u$ on ${\mathbb{R}}^n\times {\mathbb{R}}^{\mathrm{\ell }}$, with $p$ and $q$ sent to the origins. Then the fibre $F^{-1}(q)$ is represented by the slice $\{0\}\times {\mathbb{R}}^{\mathrm{\ell }}$. By [L1], this is the embedded-submanifold structure on the fibre near $p$, so its tangent vectors are exactly the vectors of the form $(0,w)$.

Let $\pi (u,v)=u$ be the coordinate projection. Step 1.1 makes $\psi \circ F\circ {\phi }^{-1}=\pi$ near the distinguished point. By [L3], the differentials $d{\phi }_p$ and $d{\psi }_q$ are isomorphisms, and [L4] gives $d{\psi }_q\circ d{F}_p\circ d({\phi }^{-1}{)}_{\phi (p)}=d{\pi }_{\phi (p)}\mathrm{.}$ Because $d{\pi }_{\phi (p)}$ has kernel $\{0\}\times {\mathbb{R}}^{\mathrm{\ell }}$, one gets $\ker d{F}_p=d{\phi }_p^{-1}(\{0\}\times {\mathbb{R}}^{\mathrm{\ell }})\mathrm{.}$ Step 1.1 identifies the same subspace with $T_p(F^{-1}(q))$, so $T_p(F^{-1}(q))=\ker d{F}_p$.

Therefore the intrinsic tangent space of the regular level set equals $\ker d{F}_p$.