For what it's worth, a complete proof of the following fact can be found in $\S 6$ of these notes.
Theorem: For a finite degree field extension $K/F$, the following are equivalent:
(i) The trace form $(x,y) \in K^2 \mapsto \operatorname{Tr}(xy) \in F$ is a nondegenerate $F$-bilinear form.
(ii) The trace form is not identically zero.
(iii) The extension $K/F$ is separable.
The specific answer to the OP's question appears in there, but here it is: always be careful to read the fine print when dealing with inseparable extensions. In this case, it turns out that when $K/F$ is inseparable, the trace from $K$ down to $F$ of $x \in K$ comes out as a power of $p$ (the characteristic of $F$) times the more familiar sum of Galois conjugates. But in characteristic $p$ multiplying something by a power of $p$ is the same as multiplying it by zero...so the trace is identically zero in inseparable extensions.
[Note also that the proof of another part of the theorem makes use of the Primitive Element Theorem, which is currently to be found in $\S 7$ of the notes, i.e., the following section. This is an obvious mistake which will have to be remedied at some point. There are plenty of other issues with these notes, which are still quite rough and incomplete.]
Denote scalar product of vectors $v,u$ by $(v,u)$, norm of vector $v$ by $\|v\|=\sqrt{(v,v)}$.
Lemma 1. Let $A$ be a symmetric positive operator on $\mathbb{R}^n$, $f\in \mathbb{R}^n$ be a vector. Then $(Af,f)\cdot (A^{-1} f,f)\geq \|f\|^4$.
Proof. Let $A=B^2$, where $B=\sqrt{A}$ is positive. Then $(Af,f)=(B^2f,f)=(Bf,Bf)=\| Bf\|^2$, $(A^{-1}f,f)=\|B^{-1}f\|^2$ and we have to prove $\|Bf\|\cdot \|B^{-1} f\|\geq \|f\|^2$, but, by the Cauchy-Schwarz inequality, $\|Bf\|\cdot \|B^{-1} f\| \geq (Bf, B^{-1} f)=(f,f)=\|f\|^2$, as desired.
Lemma 2. If for some positive symmetric matrix $A$ diagonal elements are equal to 1, then for $A^{-1}$ diagonal elements are not less then 1.
Proof. Apply Lemma 1 to basis vectors.
Now assume that $U:=V^TSV={\rm diag}(c_1,\dots,c_n)$. Then ${\rm tr}\, V^T SV=\sum c_i$,
$$
{\rm tr}\, S={\rm tr}\, V^TS(V^T)^{-1}={\rm tr}\, V^TSV (V^T V)^{-1}={\rm tr}\, U F^{-1},
$$
where $F=V^TV$ is a symmetric positive matrix with unit diagonal elements. So, by Lemma 2 diagonal elements $w_1,\dots,w_n$ of $F^{-1}$ are not less than $1$ and so ${\rm tr}\, S=\sum c_i w_i\geq \sum c_i={\rm tr}\, U$.
Best Answer
This is a purely local issue. Let $V$ be a finite-dimensional real inner product space with inner product $\langle -, - \rangle$. Then endomorphisms $T : V \to V$ can be naturally identified with bilinear forms on $V$ via the identification $T \mapsto \langle -, T(-) \rangle$. The inverse identification exists thanks to the "Riesz representation theorem" (trivial in this setting). In particular, the trace of a bilinear form can be identified with the trace of the corresponding endomorphism, and so is well-defined up to orthogonal change of coordinates.
Another way of saying this is as follows. You are correct that bilinear forms $V \times V \to \mathbb{R}$ don't have a well-defined notion of trace for $V$ only a real vector space; what has a well-defined notion of trace is an endomorphism $V \to V$, and this is because we can identify endomorphisms with elements of $V \otimes V^{\ast}$, and the dual pairing gives a distinguished map $V \otimes V^{\ast} \to \mathbb{R}$. Because one does not need to make any choices to define this map, it is automatically invariant under change of coordinates.
Bilinear forms, on the other hand, are elements of $V^{\ast} \otimes V^{\ast}$, and no analogue of the dual pairing exists here in general. However, if $V$ is an inner product space, the inner product gives a distinguished isomorphism $V \simeq V^{\ast}$ sending $v \in V$ to $\langle -, v \rangle$ and then the identification above is possible.