Let's rewrite the equation in the following way: $$ \dot{\rho}(t) = \hat{\mathcal{L}}_0\rho + \hat{\mathcal{L}}_{\textrm{J}}\rho = -i\left(H_{\textrm{eff}}(t)\rho(t) - \rho(t)H^{\dagger}_{\textrm{eff}}(t)\right) + \sum_n C_n \rho(t) C_n^{\dagger}\,, $$ where $$ H_{\textrm{eff}}(t) = H(t) - i C_n^{\dagger}C_n\,, $$ is an effective non-Hermitian Hamiltonian. As I'll show below, in this form, we can interpret the time-evolution in the following way. The "free" time-evolution governed by the first term corresponds to the standard unitary time-evolution due to the system Hamiltonian combined with an effective decay evolution "out of" the "current" state. Then, the second term is the jump term, controlling the quantum jumps into different channels.
For convenience, in what follows, we'll assume that $H(t)$ is time-independent.
To see how this interpretation works, let's move into an interaction picture by defining $$ \tilde{\rho}(t) = e^{-\mathcal{L}_0t}\rho(t)e^{\mathcal{L}_0t}\,, $$ in which case the master equation becomes $$ \frac{d\tilde{\rho}}{dt}= e^{-\mathcal{L}_0t} \hat{\mathcal{L}}_{\textrm{J}} e^{\mathcal{L}_0t} \tilde{\rho}\,. $$ Integrating this equation and plugging it into itself an infinite number of times yields a Dyson-like series, given by \begin{align} \tilde{\rho}(t)= \tilde{\rho}\left( t_{0}\right) +\sum_{n=1}^{\infty }\int_{t_{0}}^{t}dt_{n}\int_{t_{0}}^{t_{n}}dt_{n-1}\cdots \int_{t_{0}}^{t_{3}}dt_{2}\int_{t_{0}}^{t_{2}}dt_{1} e^{-\hat{\mathcal{L}}_{0}t_{n}}\hat{\mathcal{L}}_{J}e^{\hat{% \mathcal{L}}_{0}\left( t_{n}-t_{n-1}\right) }\hat{\mathcal{L}}_{J}\cdots \hat{\mathcal{L}}_{J}e^{\hat{\mathcal{L}}_{0}\left( t_{2}-t_{1}\right) }\hat{\mathcal{L}}_{J}e^{\hat{\mathcal{L}}_{0}t_{1}}\tilde{\rho}\left( t_{0}\right)\,, \end{align} and after moving out of the interaction picture, it becomes \begin{align} \hat{\rho}\left( t\right) &=e^{\hat{\mathcal{L}}_{0}\left( t-t_{0}\right) }% \hat{\rho}\left( t_{0}\right) +\sum_{n=1}^{\infty }\sum_{j_{1},\dots ,j_{n}}\int_{t_{0}}^{t}dt_{n}\int_{t_{0}}^{t_{n}}dt_{n-1}\cdots \int_{t_{0}}^{t_{3}}dt_{2}\int_{t_{0}}^{t_{2}}dt_{1}\\ &\quad\quad\mbox{}\times e^{\hat{\mathcal{L}}_{0}\left( t-t_{n}\right) }\hat{\mathcal{L}}% _{j_{n}}e^{\hat{\mathcal{L}}_{0}\left( t_{n}-t_{n-1}\right) }\hat{\mathcal{L}% }_{j_{n-1}}\cdots \hat{\mathcal{L}}_{j_{2}}e^{\hat{\mathcal{L}}_{0}\left( t_{2}-t_{1}\right) }\hat{\mathcal{L}}_{j_{1}}e^{\hat{\mathcal{L}}_{0}\left( t_{1}-t_{0}\right) }\hat{\rho}\left( t_{0}\right)\,. \end{align} The very first term represents the "free" time-evolution according to the effective Hamiltonian. We can interpret each other term in the following way.
Consider the expression inside all the integrals, given by $$ e^{\hat{\mathcal{L}}_{0}\left( t-t_{n}\right) }\hat{\mathcal{L}}% _{j_{n}}e^{\hat{\mathcal{L}}_{0}\left( t_{n}-t_{n-1}\right) }\hat{\mathcal{L}% }_{j_{n-1}}\cdots \hat{\mathcal{L}}_{j_{2}}e^{\hat{\mathcal{L}}_{0}\left( t_{2}-t_{1}\right) }\hat{\mathcal{L}}_{j_{1}}e^{\hat{\mathcal{L}}_{0}\left( t_{1}-t_{0}\right) }\hat{\rho}\left( t_{0}\right)\,. $$ We can see that we start with the density operator at the initial time $t_0$ and then evolve according to the free evolution until time $t_1$ via $e^{\hat{\mathcal{L}}_{0}\left(t_{1}-t_{0}\right) }$. At this point, we apply the jump operator $\hat{\mathcal{L}}_{j_{1}}$, which corresponds to the action $$ \rho(t_1)\to C_{j_1}\rho(t_1)C_{j_1}^{\dagger}\,. $$ We then evolve the density matrix from here via the free evolution until time $t_1$ (via $e^{\hat{\mathcal{L}}_{0}\left(t_{2}-t_{1}\right) }$), at which point we apply the jump operator $\hat{\mathcal{L}}_{j_{2}}$. We continue this until we get to time $t$.
This evolution is called a quantum trajectory, and it is not a solution to the master equation. To get that, we have to (incoherently) sum over all the different ways that this can happen. This is why there is the outer sum $\sum_{n=1}^{\infty}$, which is a sum over the number of quantum jumps that occur in the time interval $[t_0,t]$, and this is why there are the (nested) integrals, which essentially are there to sum over all the possible times at which the jumps can occur.
To see this all a little more clearly, we can assume that the initial state is a pure state, i.e., $$ \rho(t_0) = |\psi_i\rangle\langle\psi_i |\,. $$ It is relatively straightforward to see that the actions of both the free-evolution operator and of the individual jump operators preserve the purity of the state. Then, the free time-evolution is just given by $$ |\psi(t_{n})\rangle = e^{-iH_{\textrm{eff}}(t_n-t_{n-1})}|\psi(t_{n-1})\rangle\,, $$ and the action of the jump operators is just $$ |\psi(t_{n-1})\rangle\to C_n|\psi(t_{n-1})\rangle\,. $$ It is only after we incoherently sum over the different possible numbers of jumps and times at which the jumps occur that the state necessarily becomes a density matrix.
Finally, we note the following. The eigenvalues of the effective Hamiltonian necessarily have non-positive imaginary parts, because the operator $C^{\dagger}C$ is manifestly a positive operator (so its eigenvalues are real and non-negative). Thus, the eigenvalues of the free-evolution operator $e^{-iH_{\textrm{eff}}t}$ are necessarily of the form $e^{i\alpha t - \gamma t}$, where $\gamma > 0$. Thus, the free evolution corresponds to a decay "away" from the initial state, indicating that the probability of a quantum jump out of the state is increasing with time. Indeed, without going through the details, the math shows that the cumulative probability that no jumps occur in the time interval $[t_0,t]$ is given by $1 - \langle \psi(t) | \psi(t) \rangle$, indicating that the probability of a jump occurring will generally increase in time, since the $e^{-\gamma t}$ factor will guarantee that the norm of a pure state will generally decrease over time.
For much more detail, see papers about the quantum Monte Carlo jump approach in quantum optics. One good reference is a review article by Plenio and Knight.
