Why is the sampling distribution of variance a chi-squared distribution? The statement
The sampling distribution of the sample variance is a chi-squared distribution with degree of freedom equals to $n-1$, where $n$ is the sample size (given that the random variable of interest is normally distributed).
Source
My intuition
It kinda makes intuitive sense to me 1) because a chi-square test looks like a sum of square and 2) because a Chi-squared distribution is just a sum of squared normal distribution. But still, I don't have a good understanding of it.
Question
Is the statement true? Why?
 A: [I'll assume from the discussion in your question that you're happy to accept as fact that if $Z_i, i=1,2,\ldots,k$ are independent identically distributed $N(0,1)$ random variables then $\sum_{i=1}^{k}Z_i^2\sim \chi^2_k$.]
Formally, the result you need follows from Cochran's theorem.
 (Though it can be shown in other ways)
Less formally, consider that if we knew the population mean, and estimated the variance about it (rather than about the sample mean): $s_0^2 = \frac{1}{n} \sum_{i=1}^{n}(X_i-\mu)^2$, then $s_0^2/\sigma^2 = \frac{1}{n} \sum_{i=1}^{n}\left(\frac{X_i-\mu}{\sigma}\right)^2=\frac{1}{n} \sum_{i=1}^{n}Z_i^2$,  ($Z_i=(X_i-\mu)/\sigma$) which will be $\frac{1}{n}$ times a $\chi^2_n$ random variable.
The fact that the sample mean is used, instead of the population mean ($Z_i^*=(X_i-\bar{X})/\sigma$) makes the sum of squares of deviations smaller, but in just such a way that $\sum_{i=1}^{n}(Z_i^*)^2\,\sim\chi^2_{n-1}$ (about which, see Cochran's theorem). That is, rather than $ns_0^2/\sigma^2\sim \chi^2_n$ we now have $(n-1)s^2/\sigma^2\sim\chi^2_{n-1}$.
