Terminology: matrix diagonalizable as a bilinear form If a matrix $P$ is such $P^{-1}MP$ is diagonal, we say that $P$ diagonalizes $M$ (implicitly, as the matrix of an endomorphism). 
Now, if $P^\top M P$ is diagonal, is it correct to say that $P$ diagonalizes $M$ as (the matrix of) a bilinear form? It seems correct to me but I have never read this, so I am wondering.
 A: Yes, sure that is perfectly normal terminology (for example here, or here, or here or here. Frankly I'm a little surprised you never ran across this usage.)
The thing in common between the two processes is that they take two matrices that are expressing something as a matrix (resp. linear transformation, bilinear form) and connecting it with the new matrix after a change of basis (using, resp. similarity, cogredience). Whichever equivalence relation you are using, it makes perfect sense to call the classes represented by diagonal matrices "diagonalizable."
A: The natural operation for changing variables in a quadratic form amounts to the expression $P^T H P,$ where $H$ is the Hessian matrix of second partials of the form, and $P$ is nonsingular. In matrix theory, $H$ and $P^T H P $ are called congruent. There is an evident equivalence relation. 
Sometimes the Hessian, or half the Hessian, is called the Gram matrix of the form. 
When $\det P = 1,$ then $P^T H P$ and $H$ define quadratic forms that are called equivalent. In dimension two, care is taken, but in higher dimension $\det P = \pm 1$ is generally used.
If all entries of $H$ are integers and all entries of $P$ are integers, the revised form is called equivalent; if all entries of $P$ are rational, $\det P = \pm 1,$ then the forms of $H$ and $P^T HP$ are in the same genus, although there is a condition called "without essential denominator." This definition is due to Siegel. 
Maybe an example. Let $f(x,y) = x^2 + y^2.$ The Gram matrix $G$ is just the identity. Then $$f(3x+y, -5x + y) = 34 x^2 -4xy + 2 y^2.$$
Well,
$$
\left(
\begin{array}{rr}
3 & -5 \\
1 & 1
\end{array}
\right)
\left(
\begin{array}{rr}
1 & 0 \\
0 & 1
\end{array}
\right)
\left(
\begin{array}{rr}
3 & 1 \\
-5 & 1
\end{array}
\right) =
\left(
\begin{array}{rr}
34 & -2 \\
-2 & 2
\end{array}
\right)
$$  
