Let $V=F^n$ for some field $F$. Let $\cdot$ be the dot product $(v_1,...,v_n)\cdot(w_1,...,w_n)=v_1w_1+...=v_nw_n$.
Now fix two vectors $\{v,w\}$ and assume they are linearly independent. My question is, will we necessarily have $\neg \forall x \in V[v \cdot x=0$ iff $w \cdot x=0]$?
I see that the assumption ensures $n \geq 2$. Intuitively, I want to say "yes."
To prove the above, all that is necessary is to find explicitly some vector $x$ such that one of the dot products is zero and the other nonzero.
The answer to your question is yes: if $v$ and $w$ are linearly independent, then it cannot be the case that $v \cdot x = 0 \iff w \cdot x = 0$.
To see that this is the case, it suffices to consider the linear map $T:F^n \to F^2$ defined by $$ T(v) = (v \cdot x, w \cdot x). $$ The set of $x$ satisfying $v \cdot x = 0$ has dimension $n-1$ by the rank-nullity theorem. However, if $v$ and $w$ are linearly independent, then the matrix representing the transformation $T$ with respect to the usual basis has linearly independent rows, which means that $T$ has rank $2$. It follows by the rank-nullity theorem that the set of $x$ satisfying $v \cdot x = w \cdot x = 0$ has dimension $n-2$.
Since $\{x: v \cdot x = 0\}$ and $\{x:v \cdot x = w \cdot x = 0\}$ have different dimensions, they must be distinct sets. The conclusion follows.
A similar approach:
Let $f_1(x) = v \cdot x$, $f_2(x) = w \cdot x$. Since $f_1$ is non-zero, there exists an $x \in X$ with $f_1(x) \neq 0$. Define $\alpha = f_2(x)/f_1(x)$.
By linear independence, we have $f_2 \neq \alpha f_1$. So, there exists a $y \in X$ such that $$ f_2(y) \neq \alpha f_1(y) \implies f_2(y) \neq [f_2(x)/f_1(x)]f_1(y) \implies f_1(x)f_2(y) - f_1(x)f_2(y) \neq 0 $$ We note that $y$ cannot be a multiple of $x$, and neither $x$ nor $y$ are zero. Now, consider the restriction of the map $(f_1,f_2):X \to F^2$ to the span of the vectors $\{x,y\}$. We can represent this map as $$ \begin{bmatrix} f_1(ax+by)\\f_2(ax + by) \end{bmatrix} = M \begin{bmatrix} a\\b \end{bmatrix} := \begin{bmatrix} f_1(x) & f_1(y)\\ f_2(x) & f_2(y) \end{bmatrix} \begin{bmatrix} a\\b \end{bmatrix} $$ for arbitrary coefficients $a,b \in F$. By our computation above, $\det(M) \neq 0$, so that $M$ is non-singular. Thus, we may select the $a,b$ satisfying $$ \begin{bmatrix} a\\b \end{bmatrix} = M^{-1} \begin{bmatrix} 1\\0 \end{bmatrix} $$ And find that we have $f_1(ax + by) = 1$, while $f_2(ax + by) = 0$. Thus, $ax + by \in \ker(f_2) \setminus \ker(f_1)$.