Let $H:V\times V\to \mathbb{R}$ be a bilinear form. Let $\beta=\{v_1,\dots,v_n\}$ and $\gamma=\{w_1,\dots,w_n\}$ be two ordered bases for $V$, and let $Q$ be the change of basis matrix. Let there matrix representations be $\psi_\beta(H)$ and $\psi_\gamma(H)$ respectively.
Then a theorem says that $$\psi_\gamma(H)=Q^T\psi_\beta(H) Q.$$
What's confusing me is that there is no reason $Q^T=Q^{-1}$. Hence, the bilinear form is not independent of basis; in particular $$\det(\psi_\gamma(H))=\det(\psi_\beta(H))\det(Q)^2.$$ Shouldn't properties of bilinear forms be indepdent of coordinates, just like for linear transformations?
A bilinear form in the first place is a function $H\colon\>V\times V\to{\mathbb R}$ defined on pairs $(x,y)$ of points $x$, $y\in V$, and has nothing to to with bases. The choice of a basis $\beta$ in $V$ has the effect that the information inherent in $H$ can be encoded in a matrix $\psi_\beta(H)$. If the basis $\beta$ is replaced by some other basis $\gamma$ then a new matrix $\psi_\gamma(H)$ will result. Fortunately it turns out that the two matrices $\psi_\beta(H)$ and $\psi_\gamma(H)$ are related by the simple formula you quote.
You have to accept that this formula does not look the way you had hoped for, in particular that ${\rm det}\bigl(\psi_\beta(H)\bigr)$ is not an invariant of $H$. But other things that can be read off from the matrix $\psi_\beta(H)$ are, e.g., the rank of $\psi_\beta(H)$, and an eventual positive definiteness of the quadratic form$q(x):=H(x,x)$.