Formula for the unconditional variance of the sum of observations from an autoregressive time series
For an AR(1) process (I omit any drift), the coefficient on the lag is the 1st-order correlation coefficient,

$$r_{t+1} = \rho r_t + u_{t+1}$$

So

$$r_{t,t+2} = \sum_{i=1}^2 r_{t+i} = r_{t+1} + r_{t+2} = \rho r_t + u_{t+1} + \rho r_{t+1} + u_{t+2}  $$

$$=\rho r_t + u_{t+1} + \rho \big(\rho r_t + u_{t+1}\big) + u_{t+2} = \rho(1+\rho)r_t+(1+\rho)u_{t+1} + u_{t+2}$$

and now the three components are independent. So

$${\rm Var}(r_{t,t+2}) = \rho^2(1+\rho)^2\frac {\sigma^2}{1-\rho^2}+ (1+\rho)^2\sigma^2 +\sigma^2$$

$$=\left( (1+\rho)^2\frac {\rho^2+1-\rho^2}{1-\rho^2}+1\right)\cdot \sigma^2$$

$$=\left( \frac {1+\rho}{1-\rho}+1\right)\cdot \sigma^2 = \frac {2}{1-\rho}\sigma^2$$

So we arrive at a different formula for ${\rm Var}(r_{t,t+2})$, compared to the one provided in these notes, and we will arrive also at a different expression for general $n$.

Note that my result is valid for $\rho =-1$, giving ${\rm Var}(r_{t,t+2}) = \sigma^2$ in this case since here

$$r_{t+1} = - r_t + u_{t+1}$$

and so

$$r_{t+1} + r_{t+2} = - r_t + u_{t+1}- r_{t+1} + u_{t+2} = - r_t + u_{t+1} - \big(- r_t + u_{t+1}\big) + u_{t+2} = u_{t+2}$$