Help needed with applying pseudo force I have read in my book that to apply a pseudo force, 

We make sure that the object ( our reference frame ) is accelerating and then we add the negative of it's acceleration vector to the object that we are trying to apply the pseudo force at. 

Now take a look at this picture where B and A are fixed points on a circular plate. 

I want to apply a pseudo force on A w.r.t. B . I see, that B has a centripetal force $F_{c2}$ and it's direction is clearly along $BO$. 
Obviously, according to B, A will be at rest. And B knows that on A there is a centripetal force $F_{c1}$ and thus there must be some force on A ( the pseudo one ) in the $O$ to $A$ direction which should be equal to $F_{c1}$ to make A look stationary. 
Question- Why doesn't B apply a pseudo force on A which is equal to [ $-$ $F_{c2}$ ] ( according to the definition of pseudo force in my book ) but it applies a force of  [$-$  $F_{c1}$] on A ?
 A: I think your basic confusion starts when you say 

... the object ( our reference frame ) ...

The "object" and "the reference frame" are two different things. Pseudo forces are "caused" by acceleration of the reference frame, not the acceleration of the objects themselves. 
In general, the objects can also be moving and accelerating relative to the moving and accelerating reference frame.
A: Your statement about pseudo force is not entirely correct. For example, a rotating observer at the center of a merry go round is not accelerating, but still you need to add centrifugal forces.
Does your observer at $B$ self-rotate as well? If he is not self-rotating, then observed by him, $A$ will be rotating. If he is self-rotating so that he always faces $O$, $A$ will be observed to be at rest, but then you need to add another pseudo force, the centrifugal force, which together with the pseudo force you have taken into account, will exactly balance the centripetal force.
Let the position vector of $A$ and $B$ measured by $O$ be $\vec{r}_A$ and $\vec{r}_B$, respectively. Then the acceleration of $B$ itself is $$-\omega^2 \vec{r}_B$$
So one need to first add a pseudo force
$$m_A \omega^2 \vec{r}_B$$
to $A$.
Then if $B$ is self-rotating with the same $\omega$ so that it always faces $O$, then a centrifugal force
$$m_A\omega^2(\vec{r}_A - \vec{r}_B)$$
is needed as well.
So you can see that the sum is
$$m_A \omega^2 \vec{r}_A$$
which just balances the real centripetal force.
