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.
