This is a very good question, it isn't made terribly clear in the book (I only have a first edition copy of Pressley's book, but I'm assuming everything regarding these definitions is basically the same).
To use Pressley's terminology, a surface is a subset $S\subset\mathbb{R}^3$ with the property that for all $P\in S$, there are open sets $U\subset\mathbb{R}^2$, $W\subset\mathbb{R}^3$ such that $U$ and $S\cap W$ are homeomorphic, and $p\in S\cap W$. These homeomorphisms $\sigma:U\to S\cap W$ we've found are called surface patches, and given a surface $S$, the collection of them is called the atlas of $S$. Pressley then defines what it means for a surface patch to be regular, and then defines a smooth surface to be one whose atlas consists of regular surface patches.
The key issue with this definition is that given a smooth surface $S$, different people might find different atlases for it, and the way things have been set up, these would need to be considered as different surfaces, which is an undesirable situation. The usual solution is to have everyone expand the atlas they chose for $S$, to include every possible regular $\sigma:U\to S\cap W$, not just the ones you each found initially which verified $S$ was a smooth surface. Everyone will all agree on what this collection is. It is called the maximal atlas for $S$, and its members are called allowable surface patches.
So, yes, essentially the definition of "allowable surface patch" is "a regular homeomorphism $\sigma:U\to S\cap W$ where $U\subset\mathbb{R}^2$, $W\subset\mathbb{R}^3$ are open sets", and in that way allowable surface patches and regular surface patches have identical definitions. It's just that very often one wants to talk about regular surface patches which might be different than the ones you found initially, and "allowable" surface patch is a terminological distinction that indicates it may not be one from your original collection.
I think you can cover it by the two surface patchs : $$\sigma_1 : (0,2\pi)\times\mathbb{R}\to\mathbb{R^3}$$ $$\sigma_1(\theta,z)=(cos\theta,sin\theta,z)$$ and $$\sigma_2 : (-\pi,\pi)\times\mathbb{R}\to\mathbb{R^3}$$ $$\sigma_2(\theta,z)=(cos\theta,sin\theta,z)$$
"covered by surface patch" means (for example) that the support of $\sigma$ ; $\sigma((0,2\pi)\times\mathbb{R})$ is exactly the surface S, if it's not the case you must cover it by at least two surface paths.
Best Answer
Your question isn't clear, but I assume you mean that for a cylinder you can cut a piece of paper into a single shape and fold (roll) it into a cylinder, but that you cannot do that for a sphere. [For the cylinder, the flat shape is a rectangle with two disks touching opposite sides of length equal to the circumference of each disk.]
Here's is the case for a cone:
Here's the case for a cube:
The reason for this is that the Gaussian curvature of the cylinder is zero everywhere, whereas for a sphere it is zero nowhere.
A surface of zero Gaussian curvature (such as a cone, cube, etc.) can be "cut and flattened out onto a plane" (and the converse). A surface with non-zero Gaussian curvature (such as a sphere, or ellipsoid) cannot.