Kolmogorov backward equations in active biological systems
UPDATED: I wrote a first answer assuming that $t&gt;0$ which got close to what's in the paper, but not quite the same. Thanks go to Karsten Theis for pointing out that $t&lt;0$ in these "backward equations". Here's a corrected explanation:

The scenario is that we have a system that can be in any of seven consecutive states (numbered 0 through 6). At some time $t&lt;0$, we want to know something about the probability of a system in each state reaching state 6 by time $t=0$, which is expressed as $F_i(t)$ where $i$ indicates the state number.

We start with a system in state 6 at time $t$. Since reversal back to state 5 is not permitted and there is no state beyond state 6, any system that is in state 6 will remain in state 6 up to (and past) time $t=0$. Thus the probability $F_6(t)=1$. In the paper, they use a Dirac $\delta$ function to indicate this, because that is necessary when $F_6(t)$ is used in the other equations. 

For the other five states, the equations do not give explicit expressions for the probabilities, but instead express the change in probability over time as a function of the probability in a differential equation. 

Let's consider a system that at time $t$ is in state $5$. During a time interval $\Delta t$, two things can happen: Either (1) the system remains in state $5$ or (2) the system switches to state $6$.

One assumption they seem to make is that the only way the probability can change is if the system changes state. So if the system remains in state 5, its probability $F_5(t+\Delta t)$ remains equal to $F_5(t)$. If it changes to state 6, then its probability increases to $F_6(t)$. The change is thus a weighted combination of these two. 

In more detail, if we consider the probability of switching from state 5 to state 6 during the interval $\Delta t$ to be $P_{56}$, we can say that our new probability is the weighted sum 

$F_5(t+\Delta t)=P_{56}F_6(t)+(1-P_{56})F_5(t)$, 

so our change in probability is

$\Delta F_5=P_{56}F_6(t)+(1-P_{56})F_5(t)-F_5(t)=P_{56}[F_6(t)-F_5(t)]$.

Our last step is to figure out what $P_{56}$ is, and it is given by the rate constant for the change from state $5$ to state $6$. The rate constant is the average frequency of the reaction, so the probability of the reaction occurring in a given time interval is the rate constant times the length of time. So we replace $P_{56}$ with $k_{-P}\Delta t$ and get

$\Delta F_5(t)=k_{-P}\Delta t[F_6(t)-F_5(t)]$.

Dividing by $\Delta t$, we have 

$\dfrac{\Delta F_5(t)}{\Delta t}=k_{-P}[F_6(t)-F_5(t)]$.

As $\Delta t$ goes to $0$, this becomes the derivative:

$\dfrac{dF_5}{dt}=k_{-P}[F_6(t)-F_5(t)]$.

You can repeat this for the other five states to generate the other equations in the paper. The key point is that even when back reactions are permitted (such as from state 1 to state 0), the probability function is only for the systems that began in the indicated state, so the rate constant for a system coming back to that state does not enter into the equation. An implicit assumption is that the system does not make two moves during the time interval $dt$. That is, it cannot move from state 0 to state 1 and back to state 0.