As Mauro ALLEGRANZA points out, your counterexample does not work, since if $A$ is a propositional variable, it can not be a tautology: just take a valuation for which it is mapped to $0$.
I will prove a slightly stronger result by induction: for a valuation $\nu$, define $\nu^{*}$ as the valuation such that, for any propositional variable $A$, $\nu^{*}(A)=1$ if, and only if, $\nu(A)=0$. Then I state that
$$\nu(\mathfrak{B})=\nu^{*}(\neg\mathfrak{B}'),$$
for any formula $\mathfrak{B}$.
Suppose, first, that the formula $\mathfrak{B}$ has no connectives between $\vee$ and $\wedge$: since we are considering only formulas with $\vee$, $\wedge$ and $\neg$ as connectives, we must have $\mathfrak{B}=\neg\cdots\neg A$, for a propositional variable $A$ and some number of negations applied to $A$. In that case, $$\mathfrak{B}'=\mathfrak{B}\quad\text{and so}\quad\neg\mathfrak{B}'=\neg\mathfrak{B},$$ and one easily sees (apply induction in the number of negations of $\mathfrak{B}$ if you must) that $\nu(\mathfrak{B})=\nu^{*}(\neg\mathfrak{B}')$.
Suppose now that the result holds for formulas with, at most, $n-1$ ($n\geq 1$) connectives in $\{\vee, \wedge\}$, and take $\mathfrak{B}$ with $n$ of these connectives; we can write
$$\mathfrak{B}=\neg\cdots\neg(\mathfrak{C}\#\mathfrak{D})=\neg^{m}(\mathfrak{C}\#\mathfrak{D}),$$
for some $\#\in\{\vee, \wedge\}$ and some number $m$ of negations (possibly zero) applied to $\mathfrak{C}\#\mathfrak{D}$. Then we have that $\mathfrak{B}'=\neg^{m}(\mathfrak{C}'\#'\mathfrak{D}')$, where $\vee'=\wedge$ and $\wedge'=\vee$. Since $\mathfrak{C}$ and $\mathfrak{D}$ have fewer connectives among $\{\vee, \wedge\}$ than $\mathfrak{B}$, the induction hypothesis applies to them and then
$$\nu^{*}(\neg\mathfrak{B}')=\nu^{*}\Big(\neg^{m+1}(\mathfrak{C}'\#'\mathfrak{D}')\Big)=\nu^{*}\Big(\neg^{m}(\neg\mathfrak{C}'\#\neg\mathfrak{D}')\Big)=-^{m}\Big(\nu^{*}(\neg\mathfrak{C}')\#\nu^{*}(\neg\mathfrak{D}')\Big)=$$
$$-^{m}\Big(\nu(\mathfrak{C})\#\nu(\mathfrak{D})\Big)=-^{m}\nu(\mathfrak{C}\#\mathfrak{D})=\nu\Big(\neg^{m}(\mathfrak{C}\#\mathfrak{D})\Big)=\nu(\mathfrak{B}),$$
where: by "$-$" I mean the complement on the two-valued Boolean algebra (so $-0=1$ and $-1=0$); and to show $\nu^{*}(\neg^{m+1}(\mathfrak{C}'\#'\mathfrak{D}'))=\nu^{*}(\neg^{m}(\neg\mathfrak{C}'\#\neg\mathfrak{D}'))$ I used De Morgan's law. This finishes our proof.
Finally, you prove the result you want to prove by using this as lemma: if $\mathfrak{B}$ is a tautology, by noticing that $\nu^{**}=\nu$, for any valuation $\nu$ one has $\nu(\neg\mathfrak{B}')=\nu^{*}(\mathfrak{B})$, which equals $1$ since $\mathfrak{B}$ is a tautology.
Reciprocally, if $\neg\mathfrak{B}'$ is a tautology, for any valuation $\nu$ one has $\nu(\mathfrak{B})=\nu^{*}(\neg\mathfrak{B}')$, which is $1$ since $\neg\mathfrak{B}'$ is a tautology.
The basic definition is that of the satisfaction relation (page 60), that means:
"intuitively, a sequence $s = (s_1, s_2, \ldots)$ satisfies a wf $\mathscr B$ if and only if, when, for each $i$, we replace all free occurrences of $x_i$ (if any) in $\mathscr B$ by a symbol representing [the objcet of the domain of the interpretation] $s_i$, the resulting proposition is true under the given interpretation."
The relation is expressed by the following symbol: $s \vDash_M \mathscr B$.
Consider a simple example in the language of arithmetic. Formula $(x_1=0)$ will be satisfied in $\mathbb N$ by the sequence $s$ such that $s(x_1)=0$.
Now, the example of page 62 consider the general case of a formula $\mathscr B(x_{i_1},\ldots,x_{i_k})$ with $k$ free variables (listed in increasing order; see the specifications of the language: page 57).
Again, consider as formula $\mathscr B (x_1,x_2)$ the simple example: $(x_1+x_2 \le 5)$.
This formula "specifies" a binary relation in $\mathbb N \times \mathbb N$, i.e. the set of all pairs of natural numbers $n,m$ such that their sum is less-or-equal to five. Thus, $(1,1)$ and $(3,2)$ will belong to that set of pairs, because in each case the two numbers of the pairs satisfy the formula.
According to the previous definition, we have that for sequence $s_1$ such that $s_1(x_1)=s_1(x_2)=1$ and $s_2$ such that $s_2(x_1)=3$ and $s_2(x_2)=2$ we have:
$s_1 \vDash_{\mathbb N} (x_1+x_2 \le 5) \text { and } s_2 \vDash_{\mathbb N} (x_1+x_2 \le 5)$.
If we call $R$ the binary relation on $\mathbb N$ such that: $R = \{ (n,m) \mid n+m \le 5 \}$, we have that
every $2$-tuple (pair) $\langle b_1, b_2 \rangle$ in the relation $R$ satisfies formula $\mathscr B(x_1,x_2)$ in the interpretation $\mathbb N$; this will be written as
$\vDash_{\mathbb N} \mathscr B[b_1,b_2]$.
Now,
what is the difference between $\mathscr B$ and $\mathscr B[b_1,\ldots,b_k]$?
The first one is a symbol in the meta-language denoting a formula with its free variables: $\mathscr B(x_{i_1},\ldots,x_{i_k})$.
The expression $\vDash_M \mathscr B[b_1,\ldots,b_k]$ instead, is an expression of the meta-language referrin to a property of the "interpreted formula": it is an "extension" of the original notion of satisfaction and means that the sequence $s$ that maps the variables $(x_{i_1},\ldots,x_{i_k})$ to a $k$-uple $(b_1,\ldots,b_k)$, i.e. where $s(x_{i_j})=b_j$, satisfies formula $\mathscr B$ in the interpretation $M$.
Best Answer
There is no flaw in your proof. The requirement in the problem that the free variables of $\mathscr{B}$ are $y_1, \ldots, y_n$ is not necessary as the statement is true for any $y_1, \ldots, y_n$, which may include variables that do not occur free in $\mathscr{B}$ and need not includes all the free variables of $\mathscr{B}$: if $y$ is not free in $\mathscr{B}$, then $\mathscr{B}$ and $(\exists y)\mathscr{B}$ are logically equivalent and hence equisatisfiable; if $y$ and $z$ both occur free in $B$, then $(\exists y)(\exists z)\mathscr{B}$ and $(\exists z)\mathscr{B}$ are equisatisfiable. Your proof deals correctly with both of these cases.