Isotropic subvarieties of $ V(x_1^2+\dots+x_n^2-t_1^2-\dots-t_n^2) $ Let $ K $  be a field, $ \operatorname{char} K = 0, $ let $ Q = x_1^2+\dots+x_n^2-t_1^2-\dots-t_n^2 $ be the totally isotropic form, then the maximal linear isotropic
subspaces of  $ V(Q) \subset K^{2n} $ are the translates of 
$ V(x_1-t_1,\dots,x_n-t_n) $ under the action of  $ O(n,n) $  and therefore  have dimension $ n. $
Question:  Do there exist isotropic homogeneous subvarieties of $ V(Q) $ 
of dimension $ n $ which are not linear subspaces ?
 A: There is a much stronger result:  Any irreducible isotropic subvariety of dimension $n$ in $K^{2n}$ (with its split inner product) is an isotropic $n$-plane.  No assumption of homogeneity is necessary.
The proof is as follows:  First, it simplifies things to make the inner product be the spit quadratic form $Q = 2\,u_iv^i = 2(u_1v^1+\cdots +u_nv^n)$.  As you have remarked, any $n$-dimensional $Q$-isotropic linear subspace is equivalent, up to $\mathrm{Iso}(Q)$, to the subspace defined by $u_1 = \cdots = u_n = 0$.  Now, if $W\subset K^{2n}$ is an $n$-dimensional variety that is isotropic (i.e., its tangent space at each smooth point is $Q$-isotropic), then, by translation, one can assume that the origin is a smooth point of $W$ and that its tangent space at the origin is $u_i = 0$.  
Suppose that $W$ were not equal to its tangent plane at the origin.  Then it osculates to some finite order $k-1\ge1$ to its tangent plane at the origin, so it osculates to order $k$ to a graph $\Gamma$ of the form
$$
u_i = F_{ij_1\cdots j_k} \,v^{j_1}\cdots v^{j_k} = f_i(v),
$$
where $F_{ij_1\cdots j_k}$ is symmetric in its last $k\ge2$ indices, and not all of these coefficients are zero (otherwise the osculation of $W$ to its tangent plane at the origin would be higher than $k-1$).  However, then the isotropic condition, which is that the quadratic differential form $2\,\mathrm{d}u_i\circ \mathrm{d}v^i$ must vanish on $TW$, implies that this quadratic differential form must also vanish on $\Gamma$ at the origin to order at least $k{-}1$.  However, on $\Gamma$, we have
$$
2\,\mathrm{d}u_i\circ \mathrm{d}v^i = \left(\frac{\partial f_i}{\partial v^j}+\frac{\partial f_j}{\partial v^i}\right)\, \mathrm{d}v^j\circ \mathrm{d}v^i,
$$ 
and this vanishes at the origin ($v=0$) to order $k{-}1$ if and only if
$$
F_{ij_1j_2\cdots j_k} + F_{j_1ij_2\cdots j_k} = 0.
$$
Thus, $F$ must be skew-symmetric in its first two indices but symmetric in its last $k$ indices.  Since $k\ge2$, it follows that $F_{ij_1\cdots j_k}$ vanishes identically, which contradicts the choice of $k$.  
Thus, $W$ must equal its tangent plane at the origin, i.e., $W$ is an isotropic $n$-plane.
