The above notation is not standard. Usually chain groups are indexed with subscripts, while cochain groups are indexed with superscripts, but I dislike indexing with superscripts, so I'm going to use subscripts across the board.
If Ci is free of rank n, C′i is the set of functions from Ci into M, which are determined by the images of the generators. In other words, C′i = Mn. This holds even if n is an infinite cardinal.
So far C′ is a sequence of modules, but it isn't a chain, because we don't have a boundary operator, i.e. homomorphisms mapping one module into the next. We're going to build such an operator now, and since this is a cochain, it will be called the coboundary. It runs in the opposite direction of the boundary. If a boundary homomorphism map Ci into Ci+1, then an induced coboundary homomorphism map C′i+1 into C′i.
Fix a homomorphism h from Ci into M, thus defining an element of C′i. The composite function bh maps Ci-1 into Ci and over to M. This becomes an element of C′i-1. This is the coboundary operator on the cochain C′. Verify that bh1+bh2 = b*(h1+h2), and so on, whence the coboundary operator is indeed a module homomorphism.
If a map b from C6 into C7 is 0, the trivial homomorphism, it induces a trivial coboundary from C′7 back to C′6. Also, the map from C′7 to C′5 is the composition of induced maps from C′7 to C′6, and from C′6 to C′5. Of course the boundary from C5 to C7 is 0, hence the coboundary from C′7 to C′5 is 0. The image of each coboundary lies in the kernel of the next, and the cochain C′ is indeed a chain complex.
Take the kernel of a coboundary operator and mod out by the image of the previous coboundary operator to obtain the cohomology modules. This is the homology of C′, or the cohomology of C with respect to M.
Let h be a particular homomorphism in E′i. Apply f′, then b (in C′), which is the same as bfh. Then apply b (in E′), followed by f′, which yields fbh. Since f is a chain map, bf = fb, hence f′ is also a chain map.
Verify that the sum of two chain maps f+g from C into E induces a chain map from E′ to C′ that is the sum of the two maps induced by f and g separately. A similar result holds for the composition of f and g as they connect three chain complexes and their corresponding cochain complexes.
A chain map is a chain map, hence f′ induces a homomorphism on the homology of E′, which happens to be the cohomology of E. In other words, f maps the homology of C into the homology of E, and the cohomology of E into the cohomology of C. The sum f+g adds the corresponding maps on cohomology, and the composition fg composes the corresponding maps on cohomology, in the reverse order.
If A is a subchain of C, restrict f to A, and f′ maps hom(E,M) into hom(A,M). This is indeed a restriction of the original chain map f′. The induced maps on homology and cohomology are similarly restricted. The theorems regarding the join and direct sum of two restrictions of C apply here, and induce the join an direct sum of the restrictions on the map from the cohomology of E into the cohomology of C. I'll leave the details to you.