How does a string pulls pulley at both the ends? 
For an Atwood's machine, why does the string exert a downwards force on the pulley that is twice its tension? I'm neither able to understand nor imagine this scenario.
How does a wrapped string with tension $T$ pull a pulley downwards with a force of $T$ at each end?
In short I want to understand that how does tension T in rope pulls pulley ( please open the link, I was not able to upload image here)?
https://linksharing.samsungcloud.com/cdbxRTLOi3d2
 A: At one level the answer is trivial- the pulley is subject to two parallel forces T, so the resultant force is 2T.
If, however, your question is exactly how is the force applied to the pulley, the answer is that it is the vertical component of the normal force integrated over the surface of the pulley.
To see that, imagine the pulley is replaced by a square. You will easily see that a normal force exists at the top left and top right corners of the square, where the rope changes direction, and that the vertical components of those normal forces add to 2T.
If you replace the square by a hexagon, you can repeat the calculation of normal forces at each of the corners where the rope changes direction, and you will see again that the vertical components add to 2T. if you like, you can repeat it for an octagon, or any polygon with an increasing number of sides, and you will find the vertical components of the normal forces at the corners always sum to 2T. The case of a circular pulley is the same arrangement taken to an infinite number of corners of infinitely short line segments.
