I assume that we require $f(0) = 0$. The proof in the general case should be similar, if one considers large times rather than small times, but I did not work out the details.
Let $\tau_x$ be the first hitting time of $x$ by $W_t$, and $\sigma_x$ be the hitting time of $x$ by $V_t$.
Suppose that such an $f$ exists. If $x > 0$, then $V_s < x$ for $s \in f([0, \tau_x))$, and hence $\sigma_x \ge f([0, \tau_x))$. On the other hand, $\sigma_x \le f(\tau_x)$. It follows that $$\sigma_x = f(\tau_x) \qquad \text{and} \qquad f(s) < f(\tau_x) \text{ for $s < \tau_x$.} $$ The above property also holds for $x < 0$, by a similar argument. Hence, $f$ necessarily preserves the order of $\tau_x$ and $\sigma_x$.
The probability of the event $$ E_n := \{\tau_{1/n} < \tau_{-1/n} \iff \sigma_{1/n} < \sigma_{-1/n}\} $$ is equal to $1/2$. The probability of the event $$ E := \{\text{$E_n$ holds for all but finitely many $n$}\} $$ is therefore $0$ (by the Kolmogorov 0-1 law it is either zero or 1, and clearly it is not 1). However, we have seen that if $f$ with the desired properties exists, then $E_n$ holds for every $n$. Thus, the probability that such a function $f$ exists is indeed zero.