The infinitesimal case eliminates the apparent tetology because we can predict where the system will be at time $dt$ in the future from $dq = \dot q dt$.
Let's start with the one dimensional case. To vary the path over an infinitesimal time, we need at least two infinitesimal time intervals where $\dot x$ is varied by $\delta\dot x_1$ over $dt_1$, and then in the opposite sense $\delta\dot x_2$ over $dt_2$. Thus it ends up at the predicted fixed position $\dot x(dt_1+dt_2)$
The infinitesimal action is then simply the Riemann sum over $dt_1$ and $dt_2$
$$L_t = L(x,\dot x ,t),\,L_{dt1} = L(x+\dot xdt_1,\dot x,t+dt_1),\, L_{\delta \dot x dt1} = L(x+(\dot x+ \delta \dot x_1)dt_1,\dot x,t+dt_1)$$
$$ ds = L_tdt_1 + (L_t + \frac {\partial L_{dt1}}{\partial x}\dot xdt_1)dt_2$$
$$ ds+ \delta ds = (L_t + \frac {\partial L_t} {\partial \dot x}\delta \dot x_1)dt_1
+ (L_{\delta\dot x dt1}
+ \frac {\partial L_{\delta \dot x dt1}} {\partial \dot x}\delta \dot x_2
+ \frac {\partial L_{\delta \dot x dt1}} {\partial x}(\dot x+\delta\dot x_1)dt_1)dt_2$$
Since the variation in position over $dt_1$ must be compensated by that over $dt_2$, then $\delta \dot x_1dt_1 = - \delta \dot x_2dt_2$, so that
$$ \begin{align*}\delta ds &= (\frac {\partial L_t} {\partial \dot x}\delta \dot x_1- \frac {\partial L_{\delta\dot x dt1}} {\partial \dot x}\delta \dot x_1
)dt_1
+ \frac {\partial L_{\delta \dot x dt1}} {\partial x}\delta\dot x_1dt_1dt_2\\
&=0\\
&= -\frac{\frac {\partial L_{\delta\dot x}} {\partial \dot x}-\frac {\partial L_t} {\partial \dot x}}{dt_2} +\frac {\partial L_{\delta \dot x dt1}} {\partial x}\\
&= -\frac {d}{dt}\frac {\partial L}{\partial \dot x} + \frac{\partial L}{\partial x}\end{align*}$$
Obviously this generalises to any number of coordinates since the infinitesimal action can be varied independently for each $q_i$.