Let $f$ be a nondegenerate symmetric bilinear form over an $n$-dimensional vector space $X$. For each nonzero vector $x \in X$, define the dual vector $$f(x,-) \in X^{\prime}, \hspace{3mm} f(x,-): y \mapsto f(x,y).$$ Now we define $L_f:X \rightarrow X^{\prime}$ by $L_f: x\mapsto f(x,-)$.
Questions. Since $L_f$ is defined for all vectors in $X$, then what does $L_f$ do with $x = 0$? The only way I understand the definition of $L_f$ is that it maps nonzero vectors to $f(x,-)$, which makes sense because $f(x,-)$ was constructed using nonzero vectors. Is it somehow implied that $L_f$ maps $0$ by $0$?
The map $L_f:X\to X'$ is linear, so it maps the zero vector of $X$ to the zero vector of $X'$, that is the zero linear map.
Another way to see this is to use linearity in $x$: For all $y\in X$, we have $f(0,y)=f(0+0,y)=f(0,y)+f(0,y)$. Substracting $f(0,y)$ on both sides yields $f(0,y)=0$.
PS. Your definition of $L_f$ works perfectly for $x=0$, I don't understand what is puzzling you.