Consider an exponential discount factor $\delta\in(0,1)$.
Similarly, consider an exponential discount $\textit{function}$: $g(t)=\delta^t$.
Then, is defining the discount $\textit{rate}$ as below a correct way of defining one? My intuition is that the discount rate represents the rate at which the discount function decays:
$\rho=\frac{\frac{-dg(t)}{dt}}{g(t)}.$
Given a discount function $g(t)$, the discount rate is the rate at which the discount function declines over time. If time is discrete, then the (possibly time-varying) discount rate is \begin{equation} \rho(t)=-\frac{g(t)-g(t-1)}{g(t)}. \end{equation} If we divide each period into $n$ equal intervals and let $n\to\infty$, then we get the discount rate in continuous time as \begin{equation} \rho(t)=-\frac{g'(t)}{g(t)}, \end{equation} which is the same as your expression.
The (possibly time-varying) discount factor is defined as \begin{equation} \delta(t)=\frac{g(t)}{g(t-1)}. \end{equation}
With exponential discounting, the discount rate is constant, i.e. $\rho(t)=\rho$ for all $t$, and the (discrete time) discount function is given by \begin{equation} g(t)=\frac{1}{(1+\rho)^t}=\delta^t \end{equation} where $\delta=\frac{1}{1+\rho}$ is the constant discount factor.
The continuous time analogs are \begin{align} g(t)&=\mathrm e^{-\rho t}\\ \delta(t)=\delta&=\mathrm e^{-\rho} \end{align} Again, these are obtained by dividing each time period into $n$ equal intervals and observing that $\lim_{n\to\infty}(1+\frac{\rho}{n})^{-nt}=\mathrm e^{-\rho t}$.
