[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 $(n-1)s^2/\sigma^2\sim\chi^2_{n-1}$.
