The answer is 0.25 to 2DP (or more accurately, 0.247219).
The key observation is that
$$\frac{1}{f(r)g(r)h(r)} = \frac{1}{4} \left( 2\sqrt{r} - \sqrt{r-1} - \sqrt{r+1} \right).$$
Two see why, we apply the difference of two squares twice:
$$\begin{aligned} & \left(2\sqrt{r} - \sqrt{r-1} - \sqrt{r+1}\right) \left(\sqrt{r-1} + 2\sqrt{r} + \sqrt{r+1}\right) \\ &= 4r - \left(\sqrt{r-1} + \sqrt{r+1}\right)^2 \\ &= 4r - (r-1) - (r+1) - 2\sqrt{r-1}\sqrt{r+1} \\ &= 2r - 2\sqrt{r^2-1}. \end{aligned}$$
Likewise:
$$\begin{aligned} & \left(2\sqrt{r} + \sqrt{r-1} - \sqrt{r+1}\right) \left(2\sqrt{r} - \sqrt{r-1} + \sqrt{r+1}\right) \\ &= 4r - \left(\sqrt{r+1} - \sqrt{r-1}\right)^2 \\ &= 2r + 2\sqrt{r^2-1}. \end{aligned}$$
Multiplying these last two terms, we have
$$\begin{aligned} & f(r)g(r)h(r) \cdot \left(2\sqrt{r} - \sqrt{r-1} - \sqrt{r+1}\right) \\ &= \left(2r - 2\sqrt{r^2-1}\right) \left(2r + 2\sqrt{r^2-1}\right) = 4. \end{aligned}$$
This proves the key observation. Then the sum to be computed is:
$$\begin{aligned} S &= \frac{1}{4} \sum_{r=1}^{2020} \left(2\sqrt{r} - \sqrt{r-1} - \sqrt{r+1}\right) \\ &= \frac{2}{4} \sum_{r=1}^{2020} \sqrt{r} - \frac{1}{4} \sum_{r=1}^{2020} \sqrt{r-1} - \frac{1}{4} \sum_{r=1}^{2020} \sqrt{r+1} \\ &= \frac{2}{4} \sum_{r=1}^{2020} \sqrt{r} - \frac{1}{4} \left(\sum_{r=1}^{2020} \sqrt{r} - \sqrt{2020}\right) - \frac{1}{4} \left(\sum_{r=1}^{2020} \sqrt{r-1} + \sqrt{2021}\right) \\ &= \frac{1}{4} \left(\sqrt{2020} + 1 - \sqrt{2021}\right). \end{aligned}$$

To see that $S = 0.25$ to two decimal places, we note that:
$$0 \le \frac{1}{4} - S = \frac{\sqrt{2021} - \sqrt{2020}}{4} = \frac{1}{4(\sqrt{2020} + \sqrt{2021})} < \frac{1}{8\sqrt{1600}} = \frac{1}{320}.$$

Thus the difference does not affect the second decimal place.