Trace of a bilinear form?

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.