Let's take it one point at a time.

> Capacitance is defined as the ratio of the charge(Q) of one of the two
> plates and the potential difference(V) across the two plates (after
> they have been charged, of course)

Capacitance is the amount of charge that can be stored on the plates per volt across the plates, or 

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

The statement "after they have been charged, of course" is not essential to the definition of capacitance. This is because capacitance is also defined as

$$C=\frac{εA}{d}$$

where $ε$ is the permittivity of the space between the plates, $A$ is the area of each plate and $d$ is the separation between the plates. Note that voltage is not part of this definition. 

> and we know that P.D (V) means that the amount of work needed to bring
> one electron from one point to another.

Actually, PD is defined as the work required, *per unit charge*, to move the charge between two points. For a capacitor with constant field $E$ between the plates, the force exerted on a charge between the plates is $QE$ and the work done to move the charge between two points separated by a distance $d$ is $QEd$. Therefore the PD is $Ed$.

> So when we divide the number of charges by the P.D, what we are
> determining is that how many charges can possibly go from one plate to
> another by consuming that PD.

No. Dividing the PD by the number of charges, or $\frac{Ed}{Q}$ makes no sense. Now if you multiplied some total amount of charge times the PD that would give you the total work required to move all those charges a distance $d$. You don't consume PD. You consume energy moving the charges. For a capacitor connected to a battery, the battery supplies that energy.

> After they have switched the sides the PD will again get to zero. for
> example (it is just an supposition, don't take it seriously) V = 2 v
> and Q = 6 then C = 3 c/v which means that the plates can accumulate 3
> coulomb of charges. As we increase the V, Q increases too since V =
> QR/t (R = resistance) and the ratio never differs unless we change the
> physique or vicinity of the plates.

It's difficult to address all of this since it is based on an erroneous understanding of capacitance and PD, which I have already addressed. Please look over my answer and then let me know if your last statement still makes sense to you.

Hope this helps.