How work done is really understood? I know that $W=F\cdot d$. I am interested in the meaning of force here i.e.
Is it a continuous force applied till displacement? like the case of pulling trolley bag till displacement $d$ or
can it be non continuous force? Like the case of hitting pool ball with $F$, then ball covers displacement $d$
The formula $W=Fd$ gives you the work performed by a force $F$ which is acting constantly over a particle as it moves over a distance $d$. The distance that the particle moves after the force stops does not figure into the calculation of work done in any way whatsoever.
Thus, when you throw a ball in the air, you apply a certain upwards force $F_\mathrm{throw}$ over a distance $d_\mathrm{throw}$ of, say about one meter, then during your interaction with the ball you perform the work $F_\mathrm{throw}d_\mathrm{throw}$. The ball then rises to a height $h$ under the action of gravity, which means that gravity performs a work $mgh$ in slowing the ball down to a stop.
With this in mind, the concept of a 'non-continuous force', like the impulsive force felt by a billiard ball when it bounces off of a hard wall, is only an approximation. In any situation like this, there is always some range of motion (often an elastic deformation of one of the bodies involved) which gets neglected, but if you want a full account of the work involved then you need to drop that approximation.
That said, though, the force involved in hitting a billiard ball with a cue is not in this 'non-continuous' class. The impact between the cue and the ball takes a finite (if short) time and covers a finite (if short) distance, so there is no problem in calculating the work performed.
In any formula, the symbols must be defined and the conditions under which the formula holds should be specified.
To me, $W=\vec F\cdot \vec d$ is shorthand for "the work done by a constant force $\vec F$ along a straight path represented by a displacement $\vec d$".
To allow a more general situation, $W=\vec F\cdot \vec d$ is special case of $$W=\int_{\mbox{specified path from A to B}} \vec F \cdot d\vec r,$$ where $\vec F$ is a force that in general varies along the specific path from A to B, and $d\vec r$ is a tiny displacement along that specific path from A to B. Think of dividing up the path into little directed segments and consider the average force being applied during each little segment. (In this formula, the object's velocity and acceleration do not matter. Further, the force applied by other objects do not matter. It's only about the specified force of interest and the path over which that force is applied.)
The "force" can be the "force applied by an external object" or it could be "the net force". One should specify.
The "force" can vary continuously and it can also vary discontinuously, or some combination of the two. (Again, think of dividing the path into little directed segments and consider the average force over that segment.)
In the example in the answer by @EmilioPisanty , part of the motion may involve the net-force due to a variable force-vector (your hand throwing [and gravity]) for part of the path, followed by the net-force due to a different force-vector (gravity alone) for the next part of the path.
In some situations, one might be interested in the "work done by a conservative force" (like a spring force or a gravitational force or an electrostatic force).
