Reasoning behind equations in analysis of electrical circuit
I'm going to try to address the mathematical details of how the equation corresponds to the circuit, rather than talking about the bigger picture of the circuit behavior, in hope of bridging the right gaps of understanding. Note that I may use some nonstandard terminology as I have not formally studied circuit analysis.


  Kirchhoff's voltage law states that the directed sum of the voltages around a circuit equal zero, but what is the reason for the terms of the equation \$ -V_0 + RI + Q/C = 0 \$ being structured as they are? Specifically, why is the voltage given a negative sign, …


Mathematically, in order to have a sum to zero, then some terms must be positive and some terms must be negative (or degenerately, they must be all zero). It doesn't actually matter whether you put minus signs in the equation, or make the variables or constants have negative values; either will work, but there will always end up being negative numbers somewhere. You should arrange them however makes the numbers you end up with convenient to work with, which usually means, for example, choosing to have \$V_0\$, \$R\$, and \$I\$ be positive numbers.

But, of course, if we arbitrarily choose the signs of the terms we will not necessarily get the right answer; not all circuits are as simple as this one where any given choice of signs will give you either a correct answer or a physically impossible one.

In this case, we seem to be using the sign convention that if, as we proceed around the loop in the direction of the current arrow (which, note, does not in general indicate the actual direction of the current, but indicates the direction of current which corresponds to positive values of the current variable \$I\$) the voltage increases (in the typical case for that component), then the term is negative, and correspondingly if the voltage decreases then the term is positive.





Now, let's consider where those specific terms come from.


  why is the resistance multiplied by the current


The ideal behavior of a resistor is described by Ohm's law, \$V = IR\$. The \$V\$ in this equation is the voltage across the resistor, so we can just drop that right into our voltage sum.


  why is the charge on the capacitor divided by the capacitance


In the same way as Ohm's law describes ideal resistors,

$$ C = \frac{Q}{V} $$

describes ideal capacitors (without having a historic scientist's name attached to it). The theory says that a capacitor will have a linear relationship between the charge on it and the voltage across it, and the constant of proportionality is known as capacitance, \$C\$. So, if we want to know the voltage, we solve for \$V\$ and get \$Q/C\$.

Physically, this occurs because the accumulation of charge opposes the further accumulation of charge, by means of an electric field concentration which, from a circuit analysis perspective, is just another voltage difference.

(Note that if the initial condition is \$Q = 0\$ then it doesn't matter which sign we give to the capacitor's term — changing the sign corresponds to reversing the capacitor in the circuit which, since it's symmetric, doesn't change anything.)

Finally, an ideal voltage source is a component that has a fixed voltage across it, so its term in the equation is just \$V_0\$, the fixed voltage. (It is negated because of the choice of signs versus current direction I mentioned earlier.) That gives us all three terms of our sum of voltages.


  why is the rate at which charge accumulates on the capacitor, \$ \dot{Q} \$, a function of the charge of the capacitor, \$ f(Q) \$?


There's a fact/equation they didn't mention explicitly: Current is the flow of charge. As one might expect for a flow of substance, current \$I\$ flowing for time \$t\$ moves an amount of charge \$Q = It\$; or in full generality for a time-varying current,

$$Q = \int I(t) \,dt $$

When you have a current flowing through a capacitor, the charge on the capacitor is exactly this integral, because, physically, the individual charges cannot pass through the capacitor, instead causing a surplus and deficit of charges on the two capacitor plates (which is what the variable \$Q\$ for a capacitor refers to).

Now, getting back to the equation: if \$Q = \int I(t) dt \$ then also \$\dot{Q} = \frac{dQ}{dt} = I(t), \$ and if we substitute that in place of \$I\$ in our circuit equation \$ -V_0 + RI + Q/C = 0 \$ we get

$$ -V_0 + R\dot{Q} + \frac{Q}{C} = 0 $$

Solve this for \$ \dot{Q} \$ and you get the last step written in your quote, 

$$\dot{Q} = \frac{V_0}{R} - \frac{Q}{RC}.$$

Then solve this differential equation to obtain a formula for \$Q(t)\$ and you've completed the problem.