Spivak Calculus on Manifolds, Theorem 5-2

Question

In the proof Theorem 5-2 of Spivak Calculus on Mannifolds how is \begin{align*} V_2\cap M=\{f(a):(a,0)\in V_1\}? \end{align*} (That $\{f(a):(a,0)\in V_1\}=\{g(a,0):(a,0)\in V_1\}$ is clear.)

Edit: Due to a comment, Theorem 5-2 proves the equivalence of the following two definitions of a $k$-dimensional regular submanifold in $\mathbb{R}^n$.

Definition 1. A subset $M\subset\mathbb{R}^n$ is a $k$-dimensional submanifold if for every point $x\in M$ the following condition is satisfied:

(M) There is an open set $U\subset \mathbb R^n$ containing $x$, an open set $V\subset\mathbb{R}^n$, and a diffeomorphism $h:U\rightarrow V$ such that \begin{align*} h(U\cap M)&=V\cap(\mathbb{R}^k\times\{0\})\\ &=\{y\in V:y^{k+1}=\dots=y^n\}. \end{align*}

Definition 2. A subset $M\subset\mathbb{R}^n$ is a $k$-dimensional submanifold if for every point $x\in M$ the following "coordinate condition" is satisfied:

(C) There is an open set $U\subset \mathbb R^n$ containing $x$, an open set $W\subset\mathbb{R}^k$, and an injective differentiable function $f:W\rightarrow\mathbb{R}^n$ satisfying:

1. $f(W)=M\cap U$.
2. $f'(y)$ has maximum rank $k$ for each $y\in W$.
3. $f^{-1}:f(W)\rightarrow W$ is continuous.

The part of the proof mentioned above is proving Definition 2 $\Rightarrow$ Definition 1.

Answer 1

Edit: As RRTT points out in the comments, the following proof is incorrect as written.

Here is a justification:

$$\begin{align} V_2 \cap M = V_2' \cap U \cap M & = (V_2' \cap U) \cap (U \cap M) \\ & = (V_2' \cap U) \cap f(W) \\ & = (V_2' \cap U) \cap \{f(a) \colon (a,0) \in V_1'\} \\ & = \{g(a,0) \in U \cap V_2' \colon (a,0) \in V_1'\} \\ & = \{g(a,0) \colon (a,0) \in V_1' \cap g^{-1}(U \cap V_2')\} \\ & = \{g(a,0) \colon (a,0) \in V_1' \cap V_1\} \\ & = \{f(a) \colon (a,0) \in V_1\}. \\ \end{align}$$