Quick answer
The reason is because, assuming the data are i.i.d. and $X_i\sim N(\mu,\sigma^2)$, and defining
\begin{eqnarray*}
\bar{X}&=&\sum^N \frac{X_i}{N}\\
S^2 &=& \sum^{N} \frac{(\bar{X}-X_i)^2}{N-1}
\end{eqnarray*}
when forming confidence intervals, the sampling distribution associated with the sample variance ($S^2$, remember, a random variable!) is a chi-square distribution ($S^2(N-1)/\sigma^2 \sim \chi^2_{n-1}$), just as the sampling distribution associated with the sample mean is a standard normal distribution ($(\bar{X}-\mu)\sqrt{n}/\sigma \sim Z(0,1)$) when you know the variance, and with a t-student when you don't ($(\bar{X}-\mu)\sqrt{n}/S \sim T_{n-1}$).
Long answer
First of all, we'll prove that $S^2(N-1)/\sigma^2$ follows a chi-square distribution with $N-1$ degrees of freedom. After that, we'll see how this proof is useful when deriving the confidence intervals for the variance, and how the chi-square distribution appears (and why it is so useful!). Let's begin.
The proof
For this, maybe you must get used to the chi-square distribution in this Wikipedia article. This distribution has only one parameter: the degrees of freedom, $\nu$, and happens to have a Moment Generating Function (MGF) given by:
\begin{equation*}
m_{\chi^2_\nu}(t)=(1-2t)^{-\nu/2}.
\end{equation*}
If we can show that the distribution of $S^2(N-1)/\sigma^2$ has a moment generating function like this one, but with $\nu=N-1$, then we have shown that $S^2(N-1)/\sigma^2$ follows a chi-square distribution with $N-1$ degrees of freedom. In order to show this, note two facts:
If we define,
\begin{equation*}
Y = \sum \frac{(X_i-\bar{X})^2}{\sigma^2} = \sum Z_i^2,
\end{equation*}
where $Z_i\sim N(0,1)$, i.e., standard normal random variables, the moment generating function of $Y$ is given by
\begin{eqnarray*}
m_Y(t) &=& \mathbb{E}[e^{tY}]\\
&=&\mathbb{E}[e^{tZ_1^2}]\times \mathbb{E}[e^{tZ_2^2}]\times ...\mathbb{E}[e^{tZ_N^2}]\\
&=&m_{Z_i^2}(t)\times m_{Z_2^2}(t)\times ...m_{Z_N^2}(t).
\end{eqnarray*}
The MGF of $Z^2$ is given by
\begin{eqnarray*}
m_{Z^2}(t) &=& \int_{-\infty}^{\infty} f(z)\exp(tz^2)dz\\
&=&(1-2t)^{-1/2},
\end{eqnarray*}
where I have used the PDF of the standard normal, $f(z)=e^{-z^2/2}/\sqrt{2\pi}$ and, hence,
\begin{equation*}
m_Y(t)=(1-2t)^{-N/2},
\end{equation*}
which implies that $Y$ follows a chi-square distribution with $N$ degrees of freedom.
If $Y_1$ and $Y_2$ are independent and each distribute as a chi-square distribution but with $\nu_1$ and $\nu_2$ degrees of freedom, then $W=Y_1+Y_2$ distributes with a chi-square distribution with $\nu_1+\nu_2$ degrees of freedom (this follows from taking the MGF of $W$; do this!).
With the above facts, note that if you multiply the sample variance by $N-1$, you obtain (after some algebra),
\begin{equation*}
(N-1)S^2 = -n(\bar{X}-\mu)+\sum(X_i-\mu)^2,
\end{equation*}
and, hence, dividing by $\sigma^2$,
\begin{equation*}
\frac{(N-1)S^2}{\sigma^2}+\frac{(\bar{X}-\mu)^2}{\sigma^2/N}=\sum \frac{(X_i-\mu)^2}{\sigma^2}.
\end{equation*}
Note that the second term in the left-side of this sum distributes as a chi-square distribution with 1 degree of freedom, and the right-hand side sum distributes as a chi-square with $N$ degrees of freedom. Therefore, $S^2(N-1)/\sigma^2$ distributes as a chi-square with $N-1$ degrees of freedom.
Calculating the Confidence Interval for the variance.
When looking for a confidence interval for the variance, you want to know the limits $L_1$ and $L_2$ in
\begin{equation*}
\mathbb{P}\left(L_1\leq \sigma^2 \leq L_2\right) = 1-\alpha.
\end{equation*}
Let's play with the inequality inside the parenthesis. First, divide by $S^2(N-1)$,
\begin{equation*}
\frac{L_1}{S^2(N-1)}\leq \frac{\sigma^2}{S^2(N-1)} \leq \frac{L_2}{S^2(N-1)}.
\end{equation*}
And then remember two things: (1) the statistic $S^2(N-1)/\sigma^2$ has a chi-squared distribution with $N-1$ degrees of freedom and (2) the variances is always greather than zero, which implies that you can invert the inequalities, because\begin{eqnarray*}
\frac{L_1}{S^2(N-1)}\leq \frac{\sigma^2}{S^2(N-1)} &\Rightarrow&
\frac{S^2(N-1)}{\sigma^2}\leq \frac{S^2(N-1)}{L_1},\\
\frac{\sigma^2}{S^2(N-1)} \leq \frac{L_2}{S^2(N-1)} &\Rightarrow&
\frac{S^2(N-1)}{L_2} \leq \frac{S^2(N-1)}{\sigma^2},\\
\end{eqnarray*}
hence, the probability we are looking for is:
\begin{equation*}
\mathbb{P}\left(\frac{S^2(N-1)}{L_2} \leq \frac{S^2(N-1)}{\sigma^2}\leq \frac{S^2(N-1)}{L_1}\right) = 1-\alpha.
\end{equation*}
Note that $S^2(N-1)/\sigma^2 \sim \chi^2(N-1)$. We want then,
\begin{eqnarray*}
\int_{\frac{S^2(N-1)}{L_2}}^{N-1}p_{\chi^2}(x)dx &=& (1-\alpha)/2\ \ \ ,\\
\int_{N-1}^{\frac{S^2(N-1)}{L_1}}p_{\chi^2}(x)dx &=& (1-\alpha)/2\ \ \,
\end{eqnarray*}
(we integrate up to $N-1$ because the expected value of a chi-squared random variable with $N-1$ degrees of freedom is $N-1$) or, equivalently,
\begin{eqnarray*}
\int_{0}^{\frac{S^2(N-1)}{L_2}}p_{\chi^2}(x)dx=\alpha/2,\\
\int_{\frac{S^2(N-1)}{L_1}}^{\infty}p_{\chi^2}(x)dx=\alpha/2.
\end{eqnarray*}
Calling $\chi^2_{\alpha/2}=\frac{S^2(N-1)}{L_2}$ and $\chi^2_{1-\alpha/2}=
\frac{S^2(N-1)}{L_1}$, where the values $\chi^2_{\alpha/2}$ and $\chi^2_{1-\alpha/2}$ can be found in chi-square tables (in computers mainly!) and solving for $L_1$ and $L_2$,
\begin{eqnarray*}
L_1 &=& \frac{S^2(N-1)}{\chi^2_{1-\alpha/2}},\\
L_2 &=& \frac{S^2(N-1)}{\chi^2_{\alpha/2}}.
\end{eqnarray*}
Hence, your confidence interval for the variance is
\begin{equation*}
C.I.=\left(\frac{S^2(N-1)}{\chi^2_{1-\alpha/2}},
\frac{S^2(N-1)}{\chi^2_{\alpha/2}}\right).
\end{equation*}
