Imagine there is a cube with length $l = (10 \pm 0.2) \; \text{cm}$. The volume $V$ would be $l^3 = 1000 \; \text{cm}^3$ and the uncertainty in $V$, $\Delta V$ would be $\frac{10.2^3 - 9.8^3}{2} = 60.008 \; \text{cm}^3$. I would then write $V = (1000 \pm 60.008) \; \text {cm}^3$.
Now, calculating the forward uncertainty $\Delta V_f$ and backward uncertainty $\Delta V_b$ yields the following: $$\Delta V_f = 10.2^3 - 10^3 = 61.208 \; \text{cm}^3$$ $$\Delta V_b = 10^3 - 9.8^3 = 58.808 \; \text{cm}^3$$
Hence, isn't $V = (1000 \pm 60.008) \; \text {cm}^3$ not a true representation of reality since the maximum value of $V$ is $(1000 + 61.208) \; \text{cm}^3$ and not $(1000 + 60.008) \; \text{cm}^3$? Shouldn't the notation for uncertainty be something like $(1000 \pm (61.208, 58.808)) \; \text{cm}^3$ instead? Wouldn't the discrepancy be extremely large for $\Delta V_f >>> \Delta V$?
You say you have a cube whose edge length is $(10 \pm 0.2)\ \mathrm{cm}.$ Did you actually mean to say that the edge length can be anywhere between $9.8$ cm exactly and $10.2$ cm exactly? That is, it is possible that the length is $9.80001$ cm or $10.19999$ cm, but it is absolutely impossible that the length is $9.79999$ cm or $10.20001$ cm?
If that is what you meant, then it may indeed make sense to say that the volume is strictly bounded by $941.142 < V < 1061.208.$ That is how we mathematically interpret the information that $9.8 < \ell < 10.2$ and $V = \ell^3.$
However, the notation $\ell = 10 \pm 0.2$ is not generally used in mathematics to signify that $9.8 < \ell < 10.2.$ Instead, if you want a compact notation for use in an exact mathematical context that signifies that $9.8 < \ell < 10.2,$ try $\lvert \ell - 10 \rvert < 0.2.$
In most mathematical contexts, for example when solving a quadratic equation, $\ell = 10 \pm 0.2$ signifies that either $\ell = 9.8$ or $\ell = 10.2,$ and that $\ell$ is not any other value in between those two possibilities.
The people who write $\ell = (10 \pm 0.2)\ \mathrm{cm}$ to signify that $\ell$ is "approximately $10$ cm but possibly up to $0.2$ cm larger or smaller" are physicists, chemists, engineers, and other practical people. Those same people also usually keep track of the significant digits in their calculations, and in $10 \pm 0.2$ you have only three significant digits (if even that many; really it should be written $10.0 \pm 0.2$). Hence, from the point of view of the kind of person who normally would write $\ell = (10.0 \pm 0.2)\ \mathrm{cm}$ to express uncertainty among more than two possible exact values, the correct answer is that $V = (1000 \pm 60)\ \mathrm{cm}^3,$ or more explicitly $V = (1.00 \pm 0.06)\times 10^3\ \mathrm{cm}^3.$
Such people do sometimes recognize different error bounds above and below the nominal value, however. For example, if $\ell = (10.0 \pm 0.5)\ \mathrm{cm},$ then $V = \left( 1.00 \begin{matrix}\scriptsize{}+0.16 \\[-1ex] \scriptsize{}-0.14\end{matrix}\right) \times 10^3\ \mathrm{cm}^3.$ If your application is such that you are writing things like $10.0 \pm 0.2$ to signify a range of uncertainty and you need to propagate an asymmetric range, you could adopt that notation. Note that in the notation $1.00 \begin{matrix}\scriptsize{}+0.16 \\[-1ex] \scriptsize{}-0.14\end{matrix}$ it is very clear which number is added for the upper bound and which is subtracted for the lower bound, unlike in the notation $1000 \pm (61.208, 58.808).$