A quadratic form in the variables $u_i$ is expressed as $u'Du$.
Matrix $T$ consists of the n characteristic vectors (of matrix $D$): $T = [v_1\quad v_2\quad ...\quad v_n]$.
The following transformation is applied to the quadratic form:
$u=Ty$, where y is a vector with n rows.
This translates into: $(Ty)'D(Ty)\quad=\quad y'T'DTy$. Defining $R\equiv T'DT$, a new quadratic form in the variables $y_i$ appears: $u'Du\ =\ y'Ry$.
I don't understand why the following is true. How can you tell that they "take the same range of values"?
Since the $u_i$ variables and the $y_i$ variables take the same range of values, the transformation does not affect the sign definiteness of the quadratic form
(This is a citation from chapter 11 of Fundamental Methods of Mathematical Economics, by Alpha C. Chiang.)
Suppose that $u' D u$ is positive for every nonzero $u$. Consider a nonzero vector $y$, and look at $y' R y$. You can rewrite this as $(y'T') D (Ty)$. And now, letting $u$ denote $Ty$, you have $u' D u$, which you know to be positive. (Why? Because the matrix $T$ is invertible, so $Ty$ cannot be the zero vector, so the first sentence tells you it's positive.)
Letting $S = T^{-1}$, you can make exactly the same argument in reverse. So if $D$ is positive definite, so is $R$, and vice versa.
The secret lemma here is that the characteristic vectors of a quadratic form always form a basis. You can see this by writing the quadratic form as $u' M u$, and then replacing $M$ with $\frac{M + M'}{2}$. Now the matrix for your form is symmetric, and Sylvester's law of inertia says that it's diagonalizable, indeed, diagonalizable via its characteristic vectors.