Elaborating on my comment, here is a negative answer to the question as asked. I claim that there is no way to assign a vector space topology $\tau_X$ to every real vector space $X$ with a nondegenerate bilinear form which simultaneously has the following three properties:

- For any $X$, the bilinear form is continuous as a map $X\times X\to\mathbb{R}$ with respect to $\tau_X$.
- If the bilinear form on $X$ is symmetric and either positive definite or negative definite, then $\tau_X$ is the usual norm topology.
- If $T:X\to Y$ is a linear injection preserving the bilinear form, then $T$ is continuous with respect to $\tau_X$ and $\tau_Y$.

In fact, this holds even if you restrict to symmetric bilinear forms; all that matters is that they are not required to be definite. To show this, let $X$ be a direct sum of countably many copies of $\mathbb{R}^2$ with bilinear form $\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$. Let $\{e_n, f_n\}$ be the standard basis of the $n$th summand of $X$. If $Y$ is the subspace spanned by the vectors $e_n+f_n$, then the bilinear form is positive definite when restricted to $Y$. In the norm topology on $Y$, the sequence $((e_n+f_n)/n)$ converges to $0$, and so by (2) and (3) this sequence must also converge to $0$ with respect to $\tau_X$. Similarly, the vectors $e_n-f_n$ span a negative definite subspace, so $((e_n-f_n)/n)$ must converge to $0$. It follows that $(e_n/n)$ and $(f_n/n)$ also converge to $0$. Now for any sequence $a_n$ of nonzero scalars, the automorphism $T$ of $X$ that sends $e_n$ to $a_ne_n$ and $f_n$ to $f_n/a_n$ preserves the bilinear form. By (3), it now follows that $(a_n e_n)$ and $(a_n f_n)$ converge to $0$ for any scalars $a_n$. In particular, this implies that $(e_n+f_n)$ must converge to $0$. But this contradicts (1), since $\langle e_n+f_n,e_n+f_n\rangle=2$ for all $n$.