= Solution
We construct a <Hermitian hyperbolic plane over a finite field> inside each successive two-dimensional piece. By the <field norm> surjectivity proved above, choose $a\in\mathbb F_{q^2}$ with $a\bar a=-1$. From two vectors in an <orthonormal basis>, form $e=v_1+av_2$. Then $B(e,e)=1+a\bar a=0$, so $e$ is a nonzero <isotropic vector>. Nondegeneracy supplies $g$ with $B(e,g)=1$; rescaling the second argument achieves this normalization. In fact $g=v_1$ already works in this two-dimensional piece. By <quadratic finite-field norm and trace surjectivity>, choose $c$ with $c+\bar c=B(g,g)$ and put $f=g-ce$. Direct expansion gives
$$
B(e,f)=1,\qquad B(f,f)=B(g,g)-c-\bar c=0.
$$
Thus $e,f$ are <linearly independent>, and their <Gram matrix> is $\begin{pmatrix}0&1\\1&0\end{pmatrix}$, which is invertible even in characteristic two. Their span is nondegenerate, so its <orthogonal complement for a sesquilinear form> is again nondegenerate. Repeat the construction there. In even dimension this exhausts $V$; in odd dimension the final one-dimensional complement has a nonzero diagonal value, which can be normalized to one using the <field norm>. Calling its vector $d$, we obtain
$$
\boxed{B(e_i,f_j)=\delta_{ij},\quad B(e_i,e_j)=B(f_i,f_j)=0,\quad B(d,d)=1,\quad B(d,e_i)=B(d,f_i)=0,}
$$
with the $d$ terms present only in odd dimension. This proves all the requested pairings and the <basis> assertion.
The printed hint needs a qualification. If $X^2-X-1$ is irreducible over $\mathbb F_q$, its two roots are conjugate, so $\zeta+\bar\zeta=1$ and $\zeta\bar\zeta=-1$; the hinted vectors then do work. They need not work when it splits. For example, over $\mathbb F_{11}$ its roots are $4,8$, fixed by conjugation in $\mathbb F_{121}$. For either choice the proposed $e$ has norm $1+\zeta^2$, respectively $6$ or $10$, rather than zero. The norm-and-trace construction above proves the assertion without that restriction.
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