two non-degenerate quadratic forms on $GF(2)^2r$

I know this:

There are two non-degenerate quadratic forms on $GF(2)^2r$. The hyperbolic form may be taken to be $Q^+(x)=x_0 x_1 + \cdots +x_{2r-2}x_{2r-1}$ ,

and the elliptic form to be

$Q^-(x)=x^2_0 +x_0x_1 +x^2_1 +x_2x_3 + \cdots +x_{2r-2}x_{2r-1}$.

and my question is:

The long answer: over any finite field $k$ of any characteristic there are exactly two isomorphism classes of non-degenerate quadratic forms of even rank $n \ge 2$, generalizing what you have listed for even rank $2r$ in characteristic 2, and to see this conceptually it seems best to use the formalism of algebraic groups.
The "split" form $Q_n^+$ is defined to be $$x_1 x_2 + \dots + x_{n-1}x_n,$$ and to define another one we first define $Q^{-}_2$ to be the norm form for the quadratic extension $k'$ of $k$ (given by homogenizing an irreducible quadratic polynomial over $k$). Note that the equation $Q^{-}_2(x,y) = 1$ has $k$-rational solution set of size $(q^2-1)/(q-1) = q+1$ for $q := |k|$, whereas $Q^{+}_2=1$ has solution set of size $q-1$, so indeed $Q^{-}_2 \not\simeq Q^{+}_2$. We define $Q^{-}_n(x_1,\dots,x_n) = Q^{-}_2(x_1,x_2) + Q^{+}_{n-2}(x_3,\dots,x_n)$ for even $n \ge 4$. Witt cancellation is valid in even rank without restriction on the characteristic, so it shows that $Q^{-}_n \not\simeq Q^{+}_n$ for all even $n \ge 2$.
For even $n \ge 2$, to show that there are no more isomorphism classes than these we first recall that the set of isomorphism classes of rank-$n$ non-degenerate quadratic spaces $(V,Q)$ over a field $K$ are in bijection with the pointed Galois cohomology set ${\rm{H}}^1(K,{\rm{O}}_n)$. Thus, the exact sequence (for any even $n$) $$1 \rightarrow {\rm{SO}}_n \rightarrow {\rm{O}}_n \rightarrow \mathbf{Z}/2\mathbf{Z} \rightarrow 1$$ of $K$-groups defines a map of pointed sets ${\rm{H}}^1(K,{\rm{O}}_n) \rightarrow {\rm{H}}^1(K,\mathbf{Z}/2\mathbf{Z})$. Since $\mathbf{Z}/2\mathbf{Z}$ is commutative, by the "twisting method" we see that the fiber through the class of $(V,Q)$ is identified with the image of ${\rm{H}}^1(K,{\rm{SO}}(Q))$ in ${\rm{H}}^1(K,{\rm{O}}(Q))$. For finite $K$ (and still assuming $n$ is even), the pointed set ${\rm{H}}^1(K,{\rm{SO}}(Q))$ vanishes, due to Lang's theorem (since ${\rm{SO}}(Q)$ is smooth and connected). Hence, for finite $k$ and even $n$ the map of sets
$${\rm{H}}^1(k,{\rm{O}}_n) \rightarrow {\rm{H}}^1(k,\mathbf{Z}/2\mathbf{Z})$$ is injective. The target has size 2, so we're done for even $n$.
The case of odd $n$ is easier to understand since over any field $K$ we have ${\rm{O}}_n = {\rm{SO}}_n \times \mu_2$ as $K$-group schemes (where projection to $\mu_2$ is the determinant) with ${\rm{SO}}_n$ smooth and connected. Lang's theorem implies that for finite $k$ the natural map $${\rm{H}}^1(k,{\rm{O}}_n) \rightarrow {\rm{H}}^1(k,\mu_2) = k^{\times}/(k^{\times})^2$$ is bijective. Thus, again the size is 2 when the characteristic is odd, and in addition to the "split" form $Q^+_n = x_0^2 + Q^+_{n-1}$ another non-degenerate one of rank $n$ is $Q^{-}_n = x_0^2 + Q^{-}_{n-1}$. (We have $Q^{-}_n \not\simeq Q^{+}_n$ due to Witt's cancellation theorem once again, with the caveat that in odd rank Witt cancellation is only valid away from characteristic 2.)
The cohomology set is trivial when $k$ has characteristic 2 (so $Q^+_n := x_0^2 + Q^+_{n-1}$ is the only example up to isomorphism of odd rank $n$ over finite fields of characteristic 2; if we make the definition $Q^{-}_n := x_0^2 + Q^{-}_{n-1}$ then it happens to be isomorphic to $Q^{+}_n$ over $k$).