Here we consider finite-dimensional vector spaces $V$ with quadratic forms over a field $\mathbb F$ where $2\neq0$. A hyperbolic plane is any space isomorphic to $\mathbb F^2$ with the quadratic form $x\mapsto x_1^2-x_2^2$, or equivalently $x\mapsto x_1x_2$. A hyperbolic space is a direct sum of hyperbolic planes. A null vector is an element $x\in V$ mapped to $0\in\mathbb F$ by the quadratic form; we write this as $x\cdot x=0$, using a dot for the bilinear form. The trivial subspace is $O=\{0\}\subseteq V$.
According to Witt's Decomposition Theorem (7.6 in Pete Clark's notes), any space can be expressed as a direct sum
$$V=K\oplus H\oplus N$$
where $K\setminus O$ has no null vectors, $H$ is hyperbolic, and $N$ has only null vectors. ($K$ is the "anisotropic kernel" of $V$, and $N$ is the "radical" of $V$.) Furthermore, this decomposition is essentially unique; if $V=K\oplus H\oplus N=K'\oplus H'\oplus N'$, then $K\cong K'$ and $H\cong H'$ and $N=N'$.
$V$ is non-degenerate when $N=O$, or equivalently
$$\forall x\in V,\;(\forall y\in V,\;x\cdot y=0)\implies x=0.$$
$V$ is anisotropic when $H=N=O$, or equivalently
$$\forall x\in V,\;x\cdot x=0\implies x=0.$$
The property without a name, $H=O$, is equivalent to
$$\forall x\in V,\;x\cdot x=0\implies(\forall y\in V,\;x\cdot y=0).$$
$V$ is totally isotropic when $K=H=O$, or equivalently
$$\forall x\in V,\;x\cdot x=0.$$
$V$ is hyperbolic when $K=N=O$. Is there any such equivalent description of this?
If the dimension of $V$ is fixed, for example $4$, then of course we can just say there exists a basis with specified dot products:
$$\exists w\in V,\;\exists x\in V,\;\exists y\in V,\;\exists z\in V,\;(w\cdot w=x\cdot x=y\cdot y=z\cdot z \\ =w\cdot y=w\cdot z=x\cdot y=x\cdot z=0\neq w\cdot x=y\cdot z).$$
