Solving 2-CNF TQBF with #SAT solver?
First of all, it is a bit misleading to say that $\mathsf{P} \subseteq \mathsf{\#P}$, since $\mathsf{P}$ is a class of decision problems, while $\mathsf{\#P}$ is a class of counting problems. However, we can think of a decision problem as a counting problem in which the answer is 0 (no) or 1 (yes), and in that sense it is certainly true that $\mathsf{P} \subseteq \mathsf{\#P}$.

The problem $\mathsf{\#SAT}$ is $\mathsf{\#P}$-complete. This means that for any problem $A \in \mathsf{\#P}$ there exists a polynomial time reduction $f$ such that for any instance $x$ of $A$, we have
$$ A(x) = \mathsf{\#SAT}(f(x)), $$
where $A(x)$ is the value of the counting problem $A$ on instance $x$.

Since $\mathsf{2QCNF} \in \mathsf{P}$, there is a trivial reduction from $\mathsf{2QCNF}$ to $\mathsf{\#SAT}$: if the $\mathsf{2QSAT}$ is satisfiable, we output the formula $x$, and otherwise we output the formula $x \land \lnot x$. This reduction runs in polynomial time since $\mathsf{2QCNF} \in \mathsf{P}$.