What is the meaning of the eigenvalues of the matrix representation of a bilinear form? Given a bilinear form $B$ on some finite-dimensional vector space $V$, we can always represent $B$ by some matrix $A$ such that $B(v,w) = [v]^TA[w]$.  Thus we could associate the eigenvalues of $A$ with $B$.  But does that have any meaning?
What, geometrically or algebraically, do the eigenvalues of the matrix representation of a bilinear form represent?
 A: Fix a bilinear form $B$ on a finite-dimensional vector space $V$, say, over a field $\Bbb F$.
Pick two bases of $V$, say, $\mathcal E$ and $\mathcal F$, and let $P$ denote the change-of-basis matrix relating them. Then, the respective matrix representations $[B]_{\mathcal E}$ and $[B]_{\mathcal F}$ of $B$ with respect to those bases are related by
$$\phantom{(\ast)} \qquad [B]_{\mathcal F} = P^\top [B]_{\mathcal E} P . \qquad (\ast)$$
In particular, taking the determinant of both sides gives
$$\phantom{(\ast\ast)} \qquad \det[B]_{\mathcal F} = (\det P)^2 \det[B]_{\mathcal E} . \qquad (\ast\ast)$$
Since the determinant of a matrix is the product of its eigenvalues and $\det P$ can take on any nonzero value in $\Bbb F$, the spectrum (set of eigenvalues) of the matrix representation $[B]_{\mathcal E}$ of $B$ in general depends on the basis $\mathcal E$ and thus does not have intrinsic (i.e., basis-independent) meaning.
That said, bilinear forms do have some invariants, and at least some of these are expressible in terms of the eigenvalues of $[B]$.
Rank The rank of a matrix is unchanged by multiplication by an invertible matrix, so the transformation rule $(\ast)$ shows that the $\operatorname{rank} [B]_{\mathcal E}$ is an invariant of $B$, and it is equal to $n := \dim V$ less the number $n_0$ of zero eigenvalues (which thus is an invariant of $B$). We have $n_0 = \dim \ker B$, where $\ker B := \{v \in V : B(v, \,\cdot\,) = 0\}$ is the kernel of $B$.
Restricting temporarily to the symmetric case, the rank is a complete invariant for symmetric bilinear form over some fields, including algebraically closed ones not of characteristic $2$.
Theorem If $B$ is a symmetric bilinear form on a finite-dimensional vector space $V$ over an algebraically closed field of characteristic not $2$, there is a basis $\mathcal E$ of $V$ for which $$B = \operatorname{diag}(\underbrace{1, \ldots, 1}_{\operatorname{rank} B}, \underbrace{0, \ldots, 0}_{n_0}) .$$
Discriminant While $(\ast\ast)$ tells us that the determinant of $[B]$ is not an invariant of $B$, if $[B]$ is invertible it also tells us that the image of $\det B$ under the canonical quotient homomorphism $\Bbb F^\times \to \Bbb F^\times / (\Bbb F^\times)^2$ (of abelian groups) is an invariant; this quantity is the discriminant of $B$. In terms of the eigenvalues of $B$, the discriminant is just the image of their product under that map. If $\Bbb F$ is algebraically closed (in fact a much weaker condition suffices), the target is the trivial group and so the discriminant contains no information. If $\Bbb F = \Bbb Q$ for example, we can identify the quotient $\Bbb F^\times / (\Bbb F^\times)^2$ with the set of squarefree nonzero integers. (Usually discriminant is applied to symmetric, nondegenerate bilinear forms, but I see no reason not to use it for nonsymmetric matrices, too.)
Again restricting temporarily to the symmetric case, over some fields the discriminant is a full invariant of a nondegenerate, symmetric bilinear form, and over others it is not. See Theorems 11 and 12 of Kaplansky's Linear Algebra and Geometry: A Second Course for details. (Thanks to rschweib for mentioning this reference in the comments.)
Over $\Bbb R$ we have a classic classification result that we can frame in terms of eigenvalues:

Sylvester's Law of Inertia Given a real, symmetric bilinear form $B$ on a finite-dimensional vector space $V$, the number $n_+$ of positive eigenvalues, then number $n_0$ of zero eigenvalues, and the number $n_-$ of negative eigenvalues (all counting multiplicity) of the matrix representation $[B]_{\mathcal E}$, are all independent of the basis $\mathcal E$ of $V$ chosen. Moreover, these are the only invariants of real, symmetric bilinear forms in the sense that, for any such form, there is some basis $\mathcal E$ for which
  $$[B]_{\mathcal E} = \operatorname{diag}(\underbrace{1, \ldots, 1}_{n_+}, \underbrace{0, \ldots, 0}_{n_0}, \underbrace{-1, \ldots, -1}_{n_-}) .$$

We say that the bilinear form $B$ is nondegenerate if $n_0 = 0$, in which case we say that it has signature $(n_+, n_-)$. The form $B$ is positive-definite iff $n_0 = n_- = 0$ and negative-definite iff $n_0 = n_+ = 0$. Geometrically, $n_+$ ($n_-$) is the dimension of the largest subspaces of $V$ on which $B$ restricts to be positive (negative) definite, and $n_0$ is the dimension of the annihilator $\ker B := \{v \in V : B(v, \cdot) = 0 \}$. We have $\Bbb R^\times / (\Bbb R^\times)^2 \cong \{\pm 1\}$, and under this identification the discriminant of a nondegenerate, real, symmetric bilinear form is just $(-1)^{n_-}$; the discriminant is a full invariant only for $n = 1$.
Finally, properties of $B$ can impose restrictions on the possible eigenvalues of $\Bbb R$ with respect to any basis. For example, if a real bilinear form $B$ is symmetric, all of its eigenvalues are real, so the existence of a nonreal eigenvalue tells us that $B$ is not symmetric (though in practice one usually knows that before knowing the eigenvalues). The converse does not hold for $n > 1$.
