(Geometric) Intuition behind Different Types of Rank 2 Tensor (Specifically Quadratic Forms)

This is essentially a follow-up to this question: Differences between a matrix and a tensor

I think I have a good intuition/idea for the change of basis for a rank-(1,1) tensor ($A\vec{v} = \vec{w}$) via diagonalization/similarity transformations/Jordan normal form, etc.

Now that I am studying the definition of symplectic and Lorentz transformations, which involve transformations of rank-(2,2) tensors (quadratic forms, $\vec{u}^T A \vec{v}$), I don't really understand the geometric intuition for why the basis of a quadratic form changes as $T^TAT$ instead of $T^{-1}AT$ like for a rank-(1,1) tensor. This is making it more difficult for me to understand the type of "invariance" that the definitions of symplectic and Lorentz transformations are supposed to imply.

Essentially: why is the change of basis for a quadratic form have the shape $T^T A T$? Is there a good geometric intuition for this/picture to have in mind?

Does this have something to do with Sylvester's Law of Inertia or the relationship between quadratic forms and symmetric bilinear forms? http://www.ucl.ac.uk/~ucahaya/ChapterV.pdf

I guess another way to phrase the question is: what is the difference in the geometric intuitions for similarity and congruence (of matrices)? http://www.maths.qmul.ac.uk/~twm/MTH6140/la26.pdf

This is especially confusing considering I was taught that they meant the same thing in 9th grade geometry -- clearly I know very little about this. Admittedly when the matrices $T$ are orthogonal (unitary), the definitions do coincide, but they don't coincide always -- for instance symplectic matrices are NOT always orthogonal, even though they are defined as changes of basis for the matrix $J$: $A^T J A = J$.

Let me first remark that the tensors you consider here are of type $(0,2)$ and not of type $(2,2)$. Initially for such a tensor you have a bilinear form $b:V\times V\to\mathbb R$ then the obvious way how a linear isomorphism $\phi:V\to V$ acts on such a form is via $(v,w)\mapsto b(\phi(v),\phi(w))$. So the conditions you are referring to just express the fact that $b(\phi(v),\phi(w))=b(v,w)$ for all $v,w$ in terms of a matrix representation of $\phi$ with respect to some basis $\mathcal B=\{v_1,\dots,v_n\}$ for $V$. You can either proceed directly by looking at the matrix $(b_{ij})$ defined by $b_{ij}:=b(v_i,v_j)$ and compute to see that you get a transpose into the picture rather than an inverse.
Moreconceptually, you can phrase things in terms of an inner product on $V$ and and orthonormal basis $\mathcal B$ for that inner product: Fixing $v\in V$, the map $w\mapsto b(v,w)$ is a linear functional on $V$, so there is a unique element $f(v)\in V$ such that $b(v,w)=\langle f(v),w\rangle$ for. A short computation then shows that the map $v\mapsto f(v)$ is linear, so $f(v)=Jv$ for some matrix $J$. (Further $J$ is symmetric is $b$ is symmetric and skew symmetric if $b$ is skew symmetric.) But then $b(v,w)=\langle Jv,w\rangle$ and $b(Av,Aw)=\langle JAv,Aw\rangle=\langle A^tJAv,w\rangle$, and these two coincide for all $v,w$ if and only if $A^tJAv=Jv$ for all $v$ and hence if and only if $A^tJA=J$.