$\newcommand{\range}{\text{image}}\newcommand{\ann}{\text{Ann}}\newcommand{\set}[1]{\{#1\}}$
Problem: Let $V$ be a finite dimensional vector space over a field $F$ and $f$ be a symmetric bilinear form on $V$. Let $W$ be a subspace of $V$. If $f$ is non-degenerate on $W$, then $V=W\oplus W^\perp$.
Below I provide a proof of this. My problem is that my proof doesn't seem to make use of the fact that $f$ is symmetric. Is the above theorem also true if we strike out the word symmetric? If yes, then one may not read my proof and just leave a comment. If not then can one check where have I used the symmetry of $f$ in the argument?
Definitions and Notations:
Given a bilinear form $f$ on a (finite dimensional) vector space $V$, we define $R_f:V\to V^*$ as $(R_fv)(u)=f(u, v)$ for all $u, v\in V$. Clearly $R_f$ is a linear map.
For a subspace $W$ of $V$, we write $f|W$ to denotes the restriciton of $f$ to $W\times W$. It can be seen that $f|W$ is a bilinear form on $W$. We define $W^\perp$ as $\set{v\in V: f(w, v)=0 \ for \ all \ w\in W}$. Also, we write $\ann W$ to denote the annihilator of $W$.
We say that $f$ is non-degenerate on a subspace $W$ of $V$ is $R_{f|W}:W\to W^*$ has rank $\dim W$.
The Purported Proof:
It is clear that $f$ is non-degenerate on $W$ if and only if $W\cap W^\perp=\set{0}$. So it suffices to show that $\dim W+\dim W^{\perp}=\dim V$. To this end, first we note the simple fact that given $v\in V$ we have \begin{equation*} v\in W^\perp\ \iff\ R_fv\in \ann W \tag{1} \end{equation*}
Let $\pi:V^*\to V^*/\ann W$ be the canonical projection map and conider the liner map $\pi\circ R_f:V\to V^*/\ann W$.
By the Rank-Nullity Theorem we have
$$ \begin{array}{rcl} \dim V &=& \dim \ker (\pi\circ R_f) + \dim \range(\pi\circ R_f)\\ \\ &\leq& \dim \ker (\pi\circ R_f) + \dim (V^*/\ann W)\\ \\ &=& \dim \ker (\pi\circ R_f) + \dim W \end{array} $$
Note that from (1) we know that $\ker (\pi\circ R_f)=W^\perp$. Therefore $\dim V\leq \dim W^\perp+\dim W$. But since $W\cap W^\perp=\set{0}$, we also have $\dim V\geq \dim W+\dim W^\perp$. So we must have $\dim V=\dim W+\dim W^\perp$ and we are done.
For symmetric or skew-symmetric forms $(u,v)=0 \implies (v,u) = 0$. But you are right, the statement does not need any of them, just decide what $\perp$ means. The nice thing is that if non-degeneracy holds for one map, it holds for the flip one ( since, if a matrix is nonsingular, then its transpose also is). Here is a quick proof to have in your toolbox ( from Milnor&Husemoller's book Symmetric bilinear forms).
Take $v \in V$. The map $W \to F$, $u \mapsto (u,v) $ is a linear functional so it must be the dot product with a unique $w$ from $W$. Hence we have $v = w + (v - w)= w + w'$. Note that $w' \in W^{\perp}$.