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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 $\gamma: [0,1] \to Q$ such that

$$\gamma(s=0)~=~q_0\qquad\text{and}\qquad \gamma(s=1)~=~q_1,$$

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} $.

References:

1. H. Goldstein, Classical Mechanics. See the first two sentences after eq. (1.47).

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:

1. H. Goldstein, Classical Mechanics. See the first two sentences after eq. (1.47).

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) Therefore, if one would like to think of a virtual displacement $\delta q$ in terms of a curve $\gamma: [0,1] \to Q$ with

$$\gamma(s=0)~=~q_0\qquad\text{and}\qquad \gamma(s=1)~=~q_1,$$

where $s\in[0,1]$ is the curve parameter, then one can not identify the curve parameter $s$ with time $t$. For instance

$$ \gamma(s) ~=~(1-s)q_0 + sq_1. $$

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} $.

References:

1. H. Goldstein, Classical Mechanics. See the first two sentences after eq. (1.47).

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 $\gamma: [0,1] \to Q$ such that

$$\gamma(s=0)~=~q_0\qquad\text{and}\qquad \gamma(s=1)~=~q_1,$$

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} $.

References:

1. H. Goldstein, Classical Mechanics. See the first two sentences after eq. (1.47).

The key 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 $t\in[t_i,t_f]$,

$$\delta t~\equiv~ 0.$$

In other words, a virtual displacement always refers to the same time $t$.

Therefore, if one would like to view the virtual displacement $\delta q$ as a homotopy $H: Q \times [0,1] \to Q$ with

$$H(q_0,s=0)~=q_0\qquad\text{and}\qquad H(q_0,s=1)~=q_1,$$

where $s\in[0,1]$ is the homotopy parameter, then one can not identify the homotopy 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} $.

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) Therefore, if one would like to think of a virtual displacement $\delta q$ in terms of a curve $\gamma: [0,1] \to Q$ with

$$\gamma(s=0)~=~q_0\qquad\text{and}\qquad \gamma(s=1)~=~q_1,$$

where $s\in[0,1]$ is the curve parameter, then one can not identify the curve parameter $s$ with time $t$. For instance

$$ \gamma(s) ~=~(1-s)q_0 + sq_1. $$

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} $.

References:

1. H. Goldstein, Classical Mechanics. See the first two sentences after eq. (1.47).