Given $G:\{0,1\}^s \rightarrow \{0, 1\}^n$ a secure PRG, how can one prove that $G'(k_1, k_2) = G(k_1) \cdot G(k_2)$ is secure ($\cdot$ means concatenation)?
In other words, I'd like to show that if there is a distinguisher for $G'$ then this implies that there exists a distinguisher for $G$.
For example, could this distinguisher be as follows?
$A(x) = \text{round}(\frac{1}{2^n} \sum_{y \in \{0,1\}^n} A(y \cdot x))$
Since this looks like homework, I'm not going to answer the question directly (and I hope others won't either), but I'll just give some hints:
You're on a good direction. If you want to prove that $G'$ is a secure PRG, then your general approach (trying to show that a distinguisher for $G'$ implies a distinguisher for $G$) is a good strategy. Keep at it.
Your particular distinguisher $A$ is not an effective distinguisher against $G$. Hint: What is the running time to compute $A(x)$?
You can probably fix up your distinguisher (to get the runtime down to something reasonable), but that's working harder than you need to. Instead, you might want to read on....
Have you heard of the notion of a "hybrid argument"? If yes, can you see any way that it might be relevant? If no, go read up on "hybrid arguments"; they are a fundamental and important proof techniques for proving indistinguishability/distinguishability.
