Maybe an example will help you. Imagine you start working at a company today and they offer you a salary of $50,000$ USD with a promise of raising it to $55,000$ USD next year. At the same date your boss is also hired with a salary of $100,000$ USD which will be raised to $110,000$ next year. Surely you will think it is unfair your boss will get a lot more next year than you will. But let's look at it from another perspective
For you
$$
\frac{\text{Salary}(\text{today} + 1\text{yr}) - \text{Salary}(\text{today})}{\text{Salary}(\text{today})} = \frac{55,000 -50,000}{50,000} = 0.1$$
For your boss
$$
\frac{\text{Salary}(\text{today} + 1\text{yr}) - \text{Salary}(\text{today})}{\text{Salary}(\text{today})} = \frac{110,000 -100,000}{100,000} = 0.1$$
Which is exactly the same number, people usually call this a relative increment, which in this case is $10\%$. So from this point of view it is the pretty fair. To make this a bit more technical call $S$ the salary, $t$ today, and $\Delta t$ make it 1 year, so that the equations above becomes
$$
\frac{S(t + \Delta t) - S(t)}{S(t)} = r \Delta t
$$
where $r$ is the example above just means an increment of $10\%/{\rm yr}$. Now let's organize this a bit
$$
\frac{1}{S(t)} \frac{S(t + \Delta t) - S(t)}{\Delta t} = r
$$
And from this, you probably will recognize the notion of derivative. If you take $\Delta t$ a very small number you will get
$$
\frac{1}{S}\frac{{\rm d}S}{{\rm d}t} = r \equiv \frac{\dot{S}}{S}
$$
Note that this is just fancy notation to represent a fractional change of the quantity $S$. In other words: how much does the quantity $S$ change in a moment, compared with this current value.
In most cases the quantity $r$ is not a constant, but depends on time. Such in your second expression, but do not loose track of the meaning there: It tells you how the relative change of quantity $c$ is affected by other factors. For example $r$ will make it grow, but $n$ will make it decrease (because it is substracting)