By and large, lack of regularity is for convienience. The "standard" CW-decomposition of a 3-dimensional lens space $L_{p,q}$ has one 0-cell, one 1-cell, one 2-cell and one 3-cell. But it's impossible to make such a simple CW-decomposition into a regular one, since $H_1 L_{p,q} \simeq \mathbb Z_p$. A regular CW-decomposition with one cell in every dimension has $H_1$ free abelian.
Of course, the lens space has a regular CW-decomposition, but it's more work and more fuss to find it. This is much like how every manifold has a triangulation but you maybe don't want to work with a triangulation. The cellular boundary "degree term" is simpler, but there's far more cells, so the benefit of having a simple degree term is killed by having a complicated chain complex.
Presumably there are spaces that have non-regular CW-decompositions and lack regular CW-decompositions. But this is very much a fussy point-set topological curiosity -- the real reason one cares about regular vs. non-regular is the one given above. I think an example of a space where there is a CW-decomposition but no regular decomposition would be the interval $[0,1]$ attach a 2-cell, where the attaching map $f : S^1 \to [0,1]$ is given by:
write $z \in S^1$ as $z=e^{i\theta}$ with $\theta \in [0,2\pi]$.
then $f(z) = (\theta/2\pi) |\sin((2\pi)^2/\theta)|$
A little argument tells you if there was a regular CW-structure then there would have to be infinitely-many cells. But then you can argue this space does not have the weak topology of such a complex. Anyhow, something like that should work.
Consider $U$ and $V$ to be semispheres, i.e. $U$ goes from the north pole to a bit past the equator, and $V$ from a bit above the equator to the south pole. But don't imagine the line to be perpendicular to the equator plane but rather contained in it. Or if you will, construct $U$ and $V$ relative to Greenwich's meridian :)
If you stare at $U$ and $V$ for a while, they are (homotopic to) disks with a cord attached to two points of the boundary. Collapsing the disks, one gets moreover a homotopy equivalence with $S^1$ in both cases.
The intersection is now a band along the equation together with the line. Via a deformation retraction, this is homotopy equivalent to just the equator - i.e. $S^1$ - and the line, which is like a figure $8$ (collapse the line).
Hence by van Kampen we have $\pi_1(X) = \pi_1(U) \ast \pi_1(V) / \langle \langle \iota_1(\omega)\iota_2(\omega)^{-1} : \omega \in \pi_1(U \cap V)\rangle\rangle$, and $\pi_1(U) = \pi_1(V) = \Bbb Z$.
The intersection $U \cap V$, as we have said, is homotopy equivalent to a figure $8$, whose fundamental group is the free group on two generators, one for each loop. Chasing the homotopies, we obtain that the generators for $\pi_1(U \cap V)$ correspond to going through the line and then to one half of each semicircle of the equator. If $a$ and $b$ are the generators of $\pi_1(U)$ and $\pi_1(V)$, then these loops correspond to $a$ and $b$ so the relation introduced is $a = b$.
Finally, this gives $\pi_1(X) = \langle a,b : a = b\rangle \simeq \Bbb Z$, as desired. Maybe a different choice of orientations changes the presentation a little bit (e.g. the relation could be $ab = 1$, which still yields the correct group).
Best Answer
Start with a 0-cell $x$.
Attach an oriented 1-cell $e$ with both endpoints identified to $x$.
Attach a 2-cell $\sigma$ with attaching map defined on its oriented boundary circle $\partial\sigma$ as follows:
ADDED LATER: To see why this CW complex $X$ is homeomorphic to the sphere with north and south pole identified, this is the case where a picture is worth a large number of words, in order to convey the intuition. Yet it can be done with intuitive rigor also, with a few carefully chosen words, which I will attempt to supply.
First note that the characteristic map of the 2-cell $f : \sigma \to X$ is a surjective continuous map and is therefore a quotient map. So, $X$ can be reconstructed by starting with the 2-dimensional disc $\sigma$ and making the same identifications on $\sigma$ as are made by the map $f$. What are these identification? They are of two different types:
So, how do we see that this is the 2-sphere with the north and south poles identified? First do just identification 1: the quotient is the 2-sphere, and the arcs $\partial\sigma_1$, $\partial\sigma_2$ each map to a longitude line connecting the north and south poles. Next do identification 2: identify the north and south poles.