It can help to add intermediate states to the truth table; there are a lot of letters from the alphabet you haven't used :-). Name the output of the first XOR gate D and add a D column to the truth table. You may want to have clearly separated "inputs", "intermediate" and "outputs". I separate them with a vertical line, and fill out the combinations for the inputs. You have three of those, that's eight combinations.
Then fill out column D, which takes A and B as inputs. Even if there were 15 more inputs, ignore them; only A and B are relevant. Work gate per gate, divide and conquer. When you've finished column D you can fill out column S in the "outputs" section, because that only depends on D and Cin, and you have those.
Repeat for the outputs of the AND gates: add E and F to the "intermediates" section, and fill them out only looking at the columns which are input for each gate.
Let's say that we want to do a good job of testing this, but without going through the entire 2^32 space of possible operands. (It is not possible for such adder to have such a bug that it only affects a single combination of operands, requiring an exhaustive search of the 2^32 space, so it is inefficient to test it that way.)
If the individual adders are working correctly, and the ripple propagation between them works correctly, then it is correct.
I would giver priority to some test cases which focus on stressing the carry rippling, since the adders have been individually tested.
My first test case would be adding 1 to 1111..1111 which causes a carry out of every bit. The result should be zero, with a carry out of the highest bit.
(Every test case should be tried over both commutations: A + B and B + A, by the way.)
The next set set of test cases would be adding 1 to various "lone zero" patterns like 011...111, 1011...11, 110111..111, ..., 1111110. The presence of a zero should "eat" the carry propagation correctly at that bit position, so that all bits in the result which are lower than that position are zero, and all higher bits are 1 (and, of course, there is no final carry out of the register).
Another set of test cases would add these "lone 1" power-of-two bit patterns to various other patterns: 000...1, 0000...10, 0000...100, ..., 1000..000. For instance, if this is added to the operand 1111.1111, then all bits from that bit position to the left should clear, and all the bits below that should be unaffected.
Next, a useful test case might be to add all of the 16 powers of two (the "lone 1" vectors), as well as zero, to each of the 65536 possible values of the opposite operand (and of course, commute and repeat).
Finally, I would repeat the above two "lone 1" tests with "lone 11": all bit patterns which have 11 embedded in 0's, in all possible positions. This way we are hitting the situations that each adder is combining two 1 bits and a carry, requiring it to produce 1 and carry out 1.
Best Answer
Boolean Algebra Approach
The equivalence of the two forms may be proven using boolean identities
Start off with the second expression $$C_{out} = A B + (A \oplus B)C_{in}$$ Using the expansion, $$A \oplus B = \bar AB + A \bar B \quad \text{ for XOR}$$ The right hand side becomes $$AB + \bar A B C_{in} + A \bar B C_{in} \tag{1}$$ Taking B common between the forst two terms, we get (Property of Boolean Algebra) $$B(A + \bar AC_{in})$$ Using the fact that addition distributes over multiplication (In boolean Algebra) $$A + BC = (A + B)(A + C) \tag{2}$$ Hence the term becomes $$B((A + \bar A)(A + C_{in})) = AB + BC_{in}$$ Plugging this back into (1) and reusing property (2) with $$B+ \bar B C_{in}$$ We get the alternate form $$AB + BC_{in} + C_{in} \tag {3}A$$ Intuitive Approach
The form (3) is true or 1, if any two among the 3 inputs are 1. This is the case since there will be a carry if and only if (iff) we add at least 2 1's.
The XOR form, on the other hand, suggests a different viewpoint. It treats the input carry differently than the other 2 inputs A and B. Effectively it is saying that the carry will be 1 iff both A and B are 1 (The AB term) or Exactly one among A and B is 1 and Cin is 1. Note that this is being unnecessarily specific and instead we could say part 2 of the condition is that Either A or B (or both) are 1 and Cin is also 1.
In boolean form this would be, $$AB + (A+B)C_{in}$$ Which is the same as (3)
The OP is advised to refer to the properties and Axioms of Boolean Algebra to prove similar equivalences.