I presume you're asking how to equip a given space with a CW-structure. There is no general procedure to do it: as Ronnie Brown mentioned in the comments on the other answer, one can hand you with a complex algebraic variety and it might as well be possible that no one knows. One usually tries to express the space into smaller and simpler looking spaces, and then obtain a CW-structure on the whole space from "patching-up" the CW-structure on those smaller spaces. For example,
Given a pair of CW-complexes $(X, A)$ with $A \subset X$ being a subcomplex, the quotient space $X/A$ inherits a natural CW-structure obtained from disjoint union of cells in $X - A$ and a new cell $e^0$ corresponding to the pinched $A$. The attaching maps $\widetilde{f}_i : \partial e_i^n \to X/A$ are obtained from composing the attaching maps $f_i : \partial e_i^n \to X$ with the quotient map $X \to X/A$. If image of the boundary of the cells under the attaching map lies inside $A$, then the cell gets pasted to the new cell $e^0$.
Given CW-complex $X, Y$, the product space $X \times Y$ inherits a CW-structure obtained from disjoint union of product of cells $e_i^m \times e_j^n$, with $e_i^m$ being a cell in $X$ and $e_j^n$ being a cell in $Y$. The attaching maps $h_{ij} : \partial(e_i^m \times e_j^n) = e_i^m \times \partial e_j^n \cup \partial e_i^m \times e_j^n \to X \times Y$ are just the product maps $\text{id} \times g_i$ and $f_i \times \text{id}$ of the attaching maps $f_i : \partial(e_i^m) \to X$ and $g_j : \partial(e_j^n) \to Y$ on each component in the union. The topology of product of cell complexes may not agree with the product topology if the original CW-complexes aren't "nice", however. This is discussed in the appendix of Hatcher's textbook.
These two operations on spaces are the most important ones one commonly uses to build complicated spaces from simpler ones. It is, thus, quite a nice property that both product and quotient of (nice) CW-complexes are CW-complexes.
For instance, your space "sphere with poles identified" is the quotient space $S^2/\{0, 1\}$. To use our first example to obtain a CW-structure on this fellow, you have to realize $(S^2, \{0, 1\})$ as a CW-pair. This is done by giving $S^2$ the following CW-structure : start from $\{0, 1\}$, then paste $1$-cell to it, and then finally a $2$-cell to it by pasting the upper half of the boundary to $[0, 1] = \{0, 1\} \cup e^1$ by the identity map and the lower half by the antipodal map. This gives your space the CW-structure $e^0 \cup e^1 \cup e^2$, where $e^2$ is pasted to the circle $e^0 \cup e^1$ via pasting the upper half of $\partial e^2$ by pinching the equatorial $S^0$ and doing the same with the lower half, but composing with the antipodal map this time (This is not equivalent to the CW-structure in the snapshot in your question, however. Also, note that as the attaching map of $e^2$ is nullhomotopic, the space is homotopy-equivalent to $S^2 \vee S^1$)
As another example, consider $S^1 \times S^1$. Using our second example, we see that the CW-structure consists of a $0$-cell $e^0 \times e^0$, a $1$-cell $e^0 \times e^1$, another $1$-cell $e^1 \times e^0$, and finally a $2$-cell $e^1 \times e^1$. The $1$-skeleton is then disjoint union of a $0$-cell and two $1$-cells attached to the $0$-cell by the constant map. It's a bit hard to see what happens to the $2$-cell $e^1 \times e^1$, but it is a nice exercise to show that the attaching map is indeed the one obtained from the word $aba^{-1}b^{-1}$.
You are right that when people draw it as a cone, they're taking a bit of license. Identifying the circle at the end of the cylinder to a single point technically does not change the "slope" of the cylinder. A more accurate depiction would be to keep the cylinder with same slope (say, horizontal), no shrinking or tapering, and just color the boundary circle at $0$ to notate that all points on the circle are the same.
However, this is hard to understand if you're new to quotients (infinitely many colored points represent a single point??), while a cone is quite easy to understand. And as far as the topology is concerned, the two options are homeomorphic.
And the tapering or sloped cone picture shows you something: points near the apex are near each other. On the cylinder (with colored end circle) that's harder to see.
And while the tapered cone picture is not isometric to a straight cylinder with its boundary circle smashed (which is why I say they are "taking license" with this picture), they are homeomorphic. If all you care is about the topology, then it is harmless to use the easier to draw and easier to understand cone.
Best Answer
If you take the example of $X := S^1$, then $SX$ is a 2-sphere. And $\Sigma X$ is obtained from $SX$ by contracting a vertical line connecting north and south poles.
The idea is that your original basepoint in $S^1$ is now an entire interval in $SX$, and you sometimes want a canonical basepoint.