I) The important fact is here that a virtual displacement $\delta$ only affects the generalized positions $q \in Q$,
$$ \delta q ~=~ q_1 - q_0. $$
It does by definition not affect the time variable $t\in[t_i,t_f]$,
$$\delta t~\equiv~ 0,$$
cf. Ref. 1. In other words, a virtual displacement always refers to the same time $t$.
II) Let us realize a virtual displacement $\delta q$ with the help of a curve $$ [0,1]~\ni~s~~\stackrel{\gamma}{\mapsto}~~ \gamma(s)~\in~Q$$ with endpoints
$$\gamma(s=0)~=~q_0\qquad\text{and}\qquad \gamma(s=1)~=~q_1,$$
and where $s\in[0,1]$ is the curve parameter. For instance, let
$$ \gamma(s) ~=~(1\!-\! s)q_0 + sq_1. $$
Then one can not identify the curve parameter $s$ with time $t$. In particular, if one writes (infinitesimally)
$$ \delta q ~=~ \frac{\partial q}{\partial s}\delta s, $$
then $\frac{\partial q}{\partial s}$ can not be identified with the generalized velocities $\dot{q}\equiv\frac{\partial q}{\partial t} $.
TL;DR: In conclusion, OP's question seems spurred by a conflation of the physical time variable $t$ and the virtual curve parameter $s$.
References:
- H. Goldstein, Classical Mechanics. See the first two sentences after eq. (1.47).