Quadratic forms are great! They are related to some pretty great stuff such as bilinear forms and the Arf invariant. Quadratic forms in general encode the so-called "quadric surfaces" such as ellipses, hyperbolic paraboloids, and so on. The principal axis theorem, also known as the spectral theorem, is one of the most important theorems in linear algebra! It is what allows us to "transform" the quadratic forms your professor mentioned.
Take a quadratic form $q: \Bbb R^n \to \Bbb R$ defined by $x \mapsto x^tAx$. Since $A$ is symmetric (or can be made symmetric pretty easily), the principal axis theorem says we may orthogonally diagonalize it! This is what eliminates any of the cross-terms such as $x_1x_2$. Going through with the orthogonal diagonalization, $x^tAx = x^tQDQ^tx = (Q^tx)^tD(Q^tx) = y^tDy$. This matrix $D$ is diagonal, and its diagonal entries are the eigenvalues of $A$.
The significance of this "transformed" quadratic form is that it is more meaningful in terms of the information it encodes. Without those pesky cross-terms, we can see exactly what the quadric surface is without the fluff. The easiest surfaces to identify are those of the form $a_1y_1^2 + a_2y_2^2 + a_3y_3^2$ since the signs of $a_1, a_2$ and $a_3$ are how we distinguish between ellipsoid, paraboloid, etc.
They are also of great use in physics when we are dealing with the inertia tensor of a rigid body. They are about one of the coolest things we learn about in first-year linear algebra!
To add to amd's comment, given a $C^2$, real-valued function $f$ of $n$ variables, and a critical point $x_0$ of the function, we can Taylor expand $f$ to second-order to discern the nature of the critical value. That is,
$$
f(x) = f(x_0) + \tfrac{1}{2}x^tHx + o(\Vert{x}\Vert^2),
$$
where $H$ is the Hessian of $f$, and it encodes all second-order partials of $f$ at the point $x_0 \in \Bbb R^n$. Since $f$ is $C^2$, the Hessian of $f$ is symmetric, and we may orthogonally diagonalize $H$ (this is "transforming" the quadratic form via a change of variables):
$$
f(y) = f(y_0) + \tfrac{1}{2}y^tDy + o(\Vert{y}\Vert^2).
$$
From $D$, we can pick off right away whether $x_0$ (equivalently, $y_0$) is a local max, min, or neither since the entries of $D$ along its diagonal are the eigenvalues of the Hessian. If $D$ has strictly positive eigenvalues, then $x_0$ is a minimum (think concave up), and if $D$ has strictly negative eigenvalues, then $x_0$ is a maximum (think concave down). If $D$ has both positive and negative eigenvalues, $x_0$ is a saddle point.
In short, this makes the classification of extrema simpler, thanks to the fact that the second-order term in the Taylor expansion of $f$ about a critical point is itself a quadratic form.
