A: $\iint\limits_{S}\mathbf{f}\cdot\mathbf{g} dS=(\iint\limits_{S}\mathbf{f}\cdot\mathbf{n} dS)(\iint\limits_{S}\mathbf{g}\cdot\mathbf{n} dS)$
B: $\iint\limits_{S}(f+g) dS=\iint\limits_{S}{f} dS +\iint\limits_{S}gdS$.
C: $\iint\limits_{S}(c{f}) dS=c\iint\limits_{S}f dS$ where c is any constant.
D: $\iint\limits_{S\cup S'}{f} dS=\iint\limits_{S}f dS+\iint\limits_{S'}f dS$
What I have deduced so far:
I think B and C are $true$ due to the same addition property of double integrals.
Your intuition for $B$ and $C$ is correct. This is not difficult to prove. Let me use a basic example to convince you that $D$ is true. Consider two disjoint subsets of $\mathbb{R}^3$. For example, take $S$ to be the unit sphere, and $S'$ to be another unit sphere in a different location in $\mathbb{R}^3$. (Notice these are very informal terms, but this is to give you a general idea of why.) It should make sense to you that if you take the surface integral over both of them ($S \cup S'$), you just do the surface integrals separately. How else would you do it? Just think of it like doing two different surfaces separately.
On the other hand, think about why $A$ cannot be true? Is this true for single integrals? Namely:
$$\int f(x)g(x)dx \neq \int f(x) dx \int g(x)dx$$
It's not even true for derivatives. Therefore, it is not true for double integrals, and thus not true for surface integrals. Try coming up with a simple counterexample. For single integrals, I would disprove the above equation by letting $f(x)=g(x)=x$.
$$\int x \times x dx \neq \int x dx \times \int x dx$$
Try to come up with another very simple example like this and disprove the statement.
Comment with questions.
Cheers.