
Re: Ch. 7.2- Contour Integration ?s
I'll have a go at answering this: The contour integral is defined as the difference between the value of the function at the end points, so:

where

.
Let's consider this for

, as Penrose does. For any
closed contour the end points

and

are coincident but for a contour around the break point (the singularity at the origin) their arguments differ by

, so the answers differ by

, so

.
Penrose's discussion on p126 in chapter 7.2 provides one way of looking at this for
closed contours -- the important results are for closed contours.
To address your questions:
(1) The deformation idea is about changing the path to other paths using holomorphic transformations -- for example in the case of

(RTR page 126), a closed contour around the origin will give a different result for the contour integral than will a closed contour that does
not include the origin. In fact, the latter will
always give the answer
zero because any two end-points chosen will have the same argument; only a closed contour around the origin will have a different argument.
(2) Results will differ by a constant -- in fact, by

in this case -- for the reasons given. The path of the contour does not matter (amazingly!) provided it is closed and has the singularity inside it. If the singularity is outside the closed contour then the result of the integral is zero. If the singularity is inside then the result is

provided the closed contour goes around the singularity once in the anti-clockwise direction (the answer is different if you go around more than once or in a different direction).
(3) Not sure about your final question fragment
"will results differ by a constant (radius of closed loop?) but differentiate to the same derivative?". As mentioned above, the radius of the closed loop is irrelevant. Also, the answer to the integration is an actual number (usually complex) -- it's a definite integral -- this means that you
cannot differentiate it again and get back to where you started.
My discussion here is not at all rigorous, and there are other things to consider when the singularities have higher orders, but I hope it helps you to see what is going on.