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.

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.