Here is an explanation via explicit equations. First suppose $q$ is odd. Since the function field extension $\mathbf{F}_q(B)/\mathbf{F}_q(C)$ has degree $2$, it is the extension gotten by adjoining to $\mathbf{F}_q(C)$ the square root of some nonsquare element $\,f\in\mathbf{F}_q(C)$. Geometrically this means that $\,f\colon C\to\mathbf{P}^1$ is a nonconstant rational function on $C$, and $B$ is the curve defined by (the equations defining the condition that $x$ is a point on $C$ and) $\,y^2=f(x)$, and then the cover $B\to C$ is given by $(x,y)\mapsto x$. Now the quadratic twist of this cover is the projection $(x,z)\mapsto x$ mapping $B
'\to C$, where $B'$ is defined by the equations $x\in C$ and $z^2=n\cdot f(x)$, with $n$ being any prescribed nonsquare in $\mathbf{F}_q$. (Different choices of nonsquares $n$ yield isomorphic covers of $C$.) Finally, pick a point $P\in C(\mathbf{F}_q)$, and let $r$ be the order of vanishing of $\,f$ at $P$. If $r$ is odd then $P$ has ramification index $2$ in both $B\to C$ and $B'\to C$, so that $P$ lies under a unique point of each of $B(\mathbf{F}_q)$ and $B'(\mathbf{F}_q)$. Now suppose $r$ is even, and let $t\colon C\to\mathbf{P}^1$ be a function over $\mathbf{F}_q$ which vanishes at $P$ to order $1$. Let $u$ be the function $\,f/t^r$, so that $u$ has neither a zero nor a pole at $P$, whence $u(P)\in\mathbf{F}_q^{\times}$. If $u(P)$ is a square in $\mathbf{F}_q^{\times}$ then $P$ has two distinct preimages in $B(\mathbf{F}_q)$ but no preimages in $B'(\mathbf{F}_q)$. If $u(P)$ is a nonsquare in $\mathbf{F}_q^{\times}$ then $P$ has two distinct preimages in $B'(\mathbf{F}_q)$ but no preimages in $B(\mathbf{F}_q)$. Therefore in every case, $P$ lies under a combined total of two points from $B(\mathbf{F}_q)$ and $B'(\mathbf{F}_q)$.

You can do the same sort of thing if $q$ is even. In that case the degree-$2$ cover is defined by $y^2+y=f(x)$, and its quadratic twist is $y^2+y=n+f(x)$, where $n\in\mathbf{F}_q(C)$ cannot be written as $u^2+u$ with $u\in\mathbf{F}_q(C)$.
[*Added later*: I implicitly assumed that the cover is separable when $q$ is even. If it is inseparable then $\#B(\mathbf{F}_q)=\#C(\mathbf{F}_q)$.]