Let $b$ be a symmetric bilinear form on a $k$-dimensional vector space $V_k(\mathbb{F}_q)$, with $char(\mathbb{F}_q)=2$ and $k$ odd. Theorem 3.15 of the book S. Ball, Finite Geometry and Combinatorial applications states that
$$
V_k(\mathbb{F}_q)=E \oplus F,
$$
where the restriction to $E$ of $b$ is an alternating form and $F$ is a non-isotropic one-dimensional subspace.
In Corollary 3.10, it says that a non-degenerate alternanting form $b$ on $V_k(\mathbb{F}_q)$ ($k=2r$ even) is, up to a choice of a basis,
$$
b(u,v)=\sum_{i=1}^r(u_{2i-1}v_i-u_{2i}v_{2i-1}).
$$
Now, in Corollary 3.16, it states that a non-degenerate symmetric bilinear form $b$ on $V_k(\mathbb{F}_q)$ ($k=2r+1$) is, up to a choice of a basis,
$$
b(u,v)=\sum_{i=1}^r(u_{2i-1}v_i+u_{2i}v_{2i-1})+u_{2r+1}v_{2r+1}.
$$
Why this last statement is true?
Best Answer
First, subtraction is the same as addition, since we work over a field of characteristic $2$.
Applying the theorem, we get $V=E\oplus\langle f\rangle$ with $b(f,f)\ne 0$ and $b|_E$ being alternating.
Then, by the previous corollary, we can choose an appropriate basis on $E$, and let $w\notin E$ be a nonzero element of $E^\perp=\{x\mid\forall e\in E:\,b(x,e)=0\}$, which should exist since $b$ is nondegenerate. (See below.)
Then we can write $w=e+\lambda f$ with some $e\in E$ and $\lambda\ne 0$, and so, using symmetricity of $b$ and characteristic $2$, $$b(w,w)\ =\ b(e,e)+2\lambda\, b(w,f)+\lambda^2\,b(f,f)\ =\ \lambda^2\,b(f,f)\ne 0$$ Since in $\Bbb F_{2^m}$ every element is square (because $x=x^{2^m}$), $b(w,w)=\alpha^2$ and we can choose the last basis vector as $\alpha^{-1}w$ to get the desired coordinate form.
Update: Finally, to prove that $b|_E$ is nondegenerate (which also implies the existence of such $w$), assume that $E\subseteq e^\perp$ for some nonzero $e$, and choose a complementary subspace $E'\subseteq E$ with $\langle e\rangle\oplus E'=E$
Then $\dim E'=2r-1$ is odd, so the alternating bilinear form $b|_{E'}$ must be degenerate, yielding another vector $e'\in E'$ perpendicular to whole $E$.
Because of nondegeneracy, $b(e,f)\ne 0$, and thus $b(e',f)=\lambda\,b(e,f)$ with some $\lambda\in F$, so $e'-\lambda e$ would be $b$-perpendicular to the whole space.