Let x be an element in M. Recall that modules are assumed unitary unless stated otherwise. Thus 1*x = x. Replace 1 with the sum of ones from the component rings, and x is the sum of entries from the respective submodules. Thus the submodules span M.
If a nontrivial sum of elements from the submodules gives 0, let M1 participate in this linear combination. Thus x1 in M1 is spanned by the remaining submodules. Multiply by 1 in R1, and x1 remains x1. However, all the elements in the other modules drop to 0, hence x1 = 0. The submodules are independent, and M is the direct product of these submodules.
If each submodule Mi is a free Ri module with rank j, then M is free over R, with rank j. Take the first basis element from each submodule and add them together to build the first basis element for M. Repeat this for basis elements 2 through j. Multiply by components of R to produce the original basis elements in the submodules, hence M is spanned. If a linear combination produces 0 then that same linear combination, multiplied by 1i also produces 0. However, the basis elements from Mi are independent, hence they must all be 0. This holds for all i from 1 to n, and the entire linear combination is 0. The constructed set is indeed a basis for M.