I know it's a bit late, but I've just got this sabe doubt. After looking around, I've come across this Microchip Doc that shows some examples.
First, we calculate \$\text{PR2}\$. From this formula,
$$ F_\text{PWM} = \dfrac{1}{(\text{PR2} + 1) \times 4 \times T_\text{OSC} \times \text{T2CKPS}} $$
we get
$$ \text{PR2} = \dfrac{1}{F_\text{PWM} \times 4 \times T_\text{OSC} \times \text{T2CKPS}} - 1 $$
where \$T_\text{OSC} = 1/F_\text{OSC}\$, and \$\text{T2CKPS}\$ is the Timer2 prescaler value (1, 4 or 16).
Therefore, if we want \$F_\text{PWM} = 20\text{kHz}\$, and choosing \$\text{T2CKPS} = 1\$, we get \$\text{PR2} = 249\$. We should choose higher values for \$\text{T2CKPS}\$ only if \$\text{PR2}\$ exceeds 8 bits (\$\text{PR2} \gt 255\$) for the given prescale.
Now we calculate the max PWM resolution for the given frequency:
$$ \text{max PWM resolution} = \log_2(\;\dfrac{F_\text{OSC}}{F_\text{PWM}}\;) $$
That gives us \$9.9658\$ bits (I know, it sounds weird, but we'll use it like that later).
Now, let's calculate the PWM duty cycle. It is specified by the 10-bit value \$\text{CCPRxL:DCxB1:DCxB0}\$, that is, \$\text{CCPRxL}\$ bits as the most significant part, and \$\text{DCxB1}\$ and \$\text{DCxB0}\$ (bits 5 and 4 of \$\text{CCPxCON}\$) the least significant bits. Let's call this value \$\text{DCxB9:DCxB0}\$, or simply \$\text{DCx}\$. (x is the CCP number)
In our case, since we have a max PWM resolution of \$9.9658\$ bits, the PWM duty cycle (that is, the value of \$\text{DCx}\$) must be a value between \$0\$ and \$2^{9.9658} - 1 = 999\$. So, if we want a duty cycle of 50%, \$\text{DCx} = 0.5 \times 999 = 499.5 \approx 500\$.
The formula given on the datasheet (also on the linked doc),
$$\text{duty cycle} = \text{DCx} \times T_\text{OSC} \times \text{T2CKPS}$$
gives us the pulse duration, in seconds. In our case, it's equal to \$25\text{ns}\$. Since \$T_\text{PWM} = 50\text{ns}\$, it's obvious that we have a 50% duty cycle.
That said, to calculate DCx in terms of duty cycle as \$r \in [0,1]\$, we do:
$$ \text{DCx} = \dfrac{r \times T_\text{PWM}}{T_\text{OSC} \times \text{T2CKPS}} = \dfrac{r \times F_\text{OSC}}{F_\text{PWM} \times \text{T2CKPS}} $$
Answering your other questions:
2) The resolution of your PWM pulse with period \$T_\text{PWM}\$ is
$$ \dfrac{T_\text{PWM}}{2^\text{max PWM res}} $$
3) Because CCPRxL, along with DCxB1 and DCxB0, determine the pulse duration. Setting CCPRxL with a higher value than \$2^\text{max PWM res} - 1\$ means a pulse duration higher than the PWM period, and therefore you'll get a flat \$V_{DD}\$ signal.
Best Answer
To step through through various duty cycles, such as incrementing 10% each period, you need to set up an interrupt on the Timer Overflow. In that interrupt services routine, you modify the PWM duty cycle register to it's new value.
To update the duty cycle based on a switch, you do the same thing. Set up an interrupt on the input pin connected to the switch and again update the duty cycle registers during that interrupt service routine.
Here is a Microchip Tutorial on Timer interrupts including some C code. And here is the Microchip Application Note on the Capture/Compare/PWM peripheral. The examples are in Assembly. But you say you've already set up the PWMs so you should already be familiar with which registers you need to modify to change the duty cycle.