Is the concatenation of PRG output still a PRG? I am puzzled by the following question, since I am not very familiar with proof methods of Pseudo-Random Generators (I only know how to use the inequality from the PRG definition to prove or disprove, but it may not work in this case?).

For any PRG $G$, if we partition an input binary string $s$ with length $n$, into substrings $s_1$ and $s_2$ with $s=s_1||s_2$ and where $|s_1|\leq|s_2|\leq|s_1|+1$, which means almost half and half, then is the following new generator $G'$ a PRG as well?
$G'(s)=G(s_1)||G(s_2)$,which means use the two sub-strings as input to $G$, and concatenate the results. Assume $|G(n)|=n+1$, and here clearly $|G'(n)|=n+2$

I think such $G'$ is a PRG but I do not know how to prove it, and I also wonder if it could be generalized like $s=s_1||s_2|| \ldots s_m$ and $G'(s)=G(s_1)||G(s_2)||\ldots G(s_m)$, is $G'(s)$ a PRG as well?
The function $G'$, as you intuited, is indeed a PRG. The standard way to establish such results is through the hybrid argument. Details follow.

Let $G_n:\{0,1\}^n\rightarrow\{0,1\}^{n+1}$ be an $\epsilon(n)$-secure PRG; that is:
  $$G_n(U_n)\approx_{\epsilon(n)}U_{n+1},~~~~~(1)$$ 
where $\approx_{\epsilon(n)}$ denotes computational distance$^*$ of $\epsilon(n)$, and $U_n$ is the uniform distribution on $\{0,1\}^n$.

Given $(1)$, we show that $G'_n$ is an $2\epsilon(n/2)$-secure PRG (for even $n$ for the moment); that is:
  $$G_{n/2}(U_{n/2})\|G_{n/2}(U_{n/2})\approx_{2\epsilon(n/2)}U_{n+2}.~~~~~(2)$$ 
This is accomplished in two steps using a hybrid (intermediate) distribution $U_{n/2+1}\|G_{n/2}(U_{n/2})$. We first show that 
  $$G_{n/2}(U_{n/2})\|G_{n/2}(U_{n/2})\approx_{\epsilon(n/2)}U_{n/2+1}\|G_{n/2}(U_{n/2}),~~~~~(3)$$
and then that
$$U_{n/2+1}\|G_{n/2}(U_{n/2})\approx_{\epsilon(n/2)}U_{n+2}.~~~~~(4)$$
$(2)$ follows by an application of the triangle inequality to $(3)$ and $(4)$. Since the arguments for $(3)$ and $(4)$ are similar, we prove just $(3)$.

Suppose for contradiction that $G_{n/2}(U_{n/2})\|G_{n/2}(U_{n/2})$ is not $\epsilon(n/2)$-close to $U_{n/2+1}\|G_{n/2}(U_{n/2})$; then we show that $G_{n/2}(U_{n/2})$ is not $\epsilon(n/2)$-close to $U_{n/2}$ contradicting $(1)$. To be precise, given a distinguisher $\mathsf{D}$ for the former, we construct a distinguisher $\mathsf{D}'$ for the latter. The construction of $\mathsf{D}'$ is straightforward: given a challenge $y\in\{0,1\}^{n/2+1}$ (which is either $G_{n/2}(U_{n/2})$ or $U_{n/2+1}$), run $\mathsf{D}$ on $y\|G_{n/2}(U_{n/2})$ and return whatever $\mathsf{D}$ returns. If $y=G_{n/2}(U_{n/2})$ then $\mathsf{D}'$ presents  $G_{n/2}(U_{n/2})\|G_{n/2}(U_{n/2})$ to $\mathsf{D}$; otherwise, it presents $U_{n/2+1}\|G_{n/2}(U_{n/2})$. Thus, $\mathsf{D}$'s ability to distinguish the distributions in $(3)$ helps $\mathsf{D}'$ distinguish the distributions in $(1)$.

A similar argument can be made for odd values of $n$, and together you get that $G'$ is a PRG. The argument can be extended for $G'(s)=G(s_1)||G(s_2)||\ldots G(s_m)$ for $s=s_1||s_2||\ldots s_m$: one has to define a $m-1$ hybrid distributions, where the $i^{th}$ hybrid replaces the first $i$ calls of $G$ with a uniformly random string; i.e., assuming $m$ divides $n$, 
  $$\underbrace{U_{n/m+1}\|\cdots\|U_{n/m+1}}_{i \text{ terms}}\|G_{n/m}(U_{n/m})\|\cdots\|G_{n/m}(U_{n/m}).$$
Also, this PRG is only $m\cdot\epsilon(n/m)$-secure.

$^*$A distribution $X=\{X_n\}$ is $\epsilon(n)$-computationally-close from another distribution $Y=\{Y_n\}$ if for every computationally-bounded distinguisher $\mathsf{D}$,
 $$\big|\Pr[\mathsf{D}(X_n)=1]-\Pr[\mathsf{D}(Y_n)=1]\big|\leq\epsilon(n).$$
In particular $X$ is computationally indistinguishable from $Y$ if $\epsilon(n)$ is negligible in $n$.