To avoid working with the map $p \mapsto dF_p$ whose codomain is space of linear maps, let's work with charts. Choose smooth charts $(V,\varphi, x^i)$ contain $p$ and $(W,\psi)$ contain $F(p)$. Denote $\hat{p} = \varphi(p)$. We actually considering the map from $\hat{V}=\varphi(V) \subseteq \mathbb{R}^m$ to space of matrices $M(m\times n,\mathbb{R})$,
$$
J: \hat{V} \rightarrow M(n\times m,\mathbb{R})
$$
defined by
$$
J : \hat{q} \longmapsto \Bigg[\frac{\partial \hat{F}^i}{\partial x^j}(\hat{q})\Bigg] \in M(n\times m,\mathbb{R}), \quad \forall \hat{q} \in \hat{V}.
$$
This map is smooth (hence continous) since each entries is a smooth functions of $\hat{q} \in \hat{V}$. By hypothesis $J(\hat{p})$ contain in an open subset of full rank matrices. Now you can argue by continuity of $J$.
Tu considers the smooth map $F : \mathbb R^3 \to \mathbb R^3, F(x,y,z) = (f(x,y,z),y,z)$ and shows that on some $U_p$ it restricts to a chart $F \mid_{U_p} : U_p \to F(U_p)$ on $\mathbb R^3$. In this special case this means nothing else than that $F \mid_{U_p}$ is diffeomorphism between open subsets of $\mathbb R^3$. The three coordinate functions of $F$ are $F_1= f$, $F_2 = p_2$, $F_3 = p_3$, where $p_i : \mathbb R^3 \to \mathbb R,p_i(x_1,x_2,x_3) = x_i$, is the projection onto the $i$-th coordinate.
But now by definition of $f$ we have have $S^2 = f^{-1}(0)$, i.e. $S^2 = \{(x,y,z) \in \mathbb R^3 \mid f(x,y,z) = 0 \}$. Therefore
$$U_p \cap S^2 = \{(x,y,z) \in U_p \mid f(x,y,z) = 0 \} = \{(x,y,z) \in U_p \mid F_1(x,y,z) = 0\} = \{(x,y,z) \in U_p \mid F(x,y,z) \in \{ 0 \} \times \mathbb R^2\} = (F \mid_{U_p})^{-1}(\{ 0 \} \times \mathbb R^2) .$$
This means that $U_p \cap S^2$ is the set of points $\xi \in U_p$ where the first coordinate of $F(\xi)$ vanishes, i.e. where $f(\xi)$ vanishes. This is what Tu expresses in the form
In this chart, the set $U_p \cap S^2$ is defined by the vanishing of the first coordinate $f$.
Note that this shows that (cf. Definition 9.1) $S^2 \subset \mathbb R^3$ is a regular submanifold of dimension $2$ because for each $p \in S^2$ we found the chart $(U_p,F \mid_{U_p})$ in the maximal atlas of $\mathbb R^3$ such that $U_p \cap S^2$ is defined by the vanishing of $n-2 = 1$ of the coordinate functions of $F \mid_{U_p}$.
Best Answer
Quick answer:
The differential of a map associate a vector $df[v]$ to any vector $[v]$, and that vector $df[v]$ depends only on $v$, not on the coordinate system.
Detailed discussion:
What is a map? A map $f:M\to \mathbb R^n$ associates to any point $x\in M$ a vector $f(x)$. The value $f(x)$ depends only on the point, not on local coordinates.
What is the differential of a map? The differential of a map is not a function from $M$ to $\mathbb R^n$. One usual way to define the differential of a function $f:M\to \mathbb R^n$ is the following:
$f$ is differentiable at $x\in M$ if there exists a linear map $L$ so that $$f(x+\epsilon v)=f(x)+\epsilon L(v)+o(\epsilon)$$
In other words, if $f$ is approximable by a linear map at first order. So, now look at $L$. Who is $L$? It is a linear map bur from where to where? The origin vector space is the tangent space of $M$ at point $x$, usually denoted $T_xM$, and the target space is the tanget space at $f(x)$. In the present case, we can canonically identity the tangent space at $f(x)$ with $\mathbb R^n$ (but if you have $f:M\to N$ you have to take $T_{f(x)}N$). So $$L\in\hom(T_xM,\mathbb R^n)$$.
The map $L$ is called the differential of $f$ at $x$ and it is usually denoted by $d_xf$.
Now, it is a general linear algebra fact that if you have two finite dimensional vector spaces $V,W$ then, for any choices of basis $B_V,B_W$ of $V$ and $W$, you can associate to any $F\in\hom(V,W)$ the matrix of $F$ in those chosen basis. The linear map $F$ exists independently on $B_V,B_W$: it is the matrix that changes when you change coordinates.
Well, for $df$ is the same: the linear map $d_xf:T_xM\to \mathbb R^n$ does not depend on coordinates, the Jacobian matrix --- which describes $d_xf$ as matrix --- does indeed depend on coordinates.