In my textbook, Laurent's theorem is presented in this way:
Let $f(z)$ be analytic in a domain containing two concentric circles $c_{1}$ and $c_{2}$with center $z_{0}$ and the annulus between them (blue in Fig. 370). Then $f(z)$ can be represented by the Laurent series $$f(z)=g(z)+h(z)=\sum\limits_{n=0}^{\infty}{a_{n}(z-z_{0})^{n}}+\sum\limits_{n=1}^{\infty}{{{b_{n}}\over{(z-z_{0})^{n}}}}$$ $$=a_{0}+a_{1}(z-z_{0})+a_{2}(z-z_{0})^{2}+\cdots$$ $$+{{b_{1}}\over{z-z_{0}}}+{{b_{2}}\over{(z-z_{0})^{2}}}+\cdots$$ consisting of nonnegative and negative powers. The coefficients of this Laurent series are given by the integrals $$a_{n}={{1}\over{2\pi i}}\oint_{c}{{{f(z^{\ast})}\over{(z^{\ast}-z_{0})^{n+1}}}}dz^{\ast}$$ and $$b_{n}={{1}\over{2\pi i}}\oint_{c}{(z^{\ast}-z_{0})^{n-1}f(z^{\ast})}dz^{\ast}$$ taken counterclockwise around any simple closed path C that lies in the annulus and encircles the inner circle, as in Fig. 370. [The variable of integration is denoted by $z^{\ast}$]
This series converges and represents $f(z)$ in the enlarged open annulus obtained from the given annulus by continuously increasing the radius of outer circle $c_{1}$ and decreasing the radius of inner circle $c_{2}$ until each of the two circles reaches a point where $f(z)$ is singular.
the author first proves the above theorem for the annulus between $c_{1}$ and $c_{2}$ and then proves the convergence of the formula $$f(z)=g(z)+h(z)=\sum\limits_{n=0}^{\infty}{a_{n}(z-z_{0})^{n}}+\sum\limits_{n=1}^{\infty}{{{b_{n}}\over{(z-z_{0})^{n}}}}$$ in the enlarged annulus in this way:
Convergence of $f(z)=g(z)+h(z)=\sum\limits_{n=0}^{\infty}{a_{n}(z-z_{0})^{n}}+\sum\limits_{n=1}^{\infty}{{{b_{n}}\over{(z-z_{0})^{n}}}}$ in the enlarged annulus.
The first series in this formula is a Taylor series [representing $g(z)$]; hence it converges in the disk D with center $z_{0}$ whose radius equals the distance of the singularity (or singularities) outside $c_{1}$ closest to $z_{0}$. Also, $g(z)$ must be singular at all points outside $c_{1}$ where $f(z)$ is singular.The second series representing $h(z)$ is a power series in $Z={{1}\over{z-z_{0}}}$. Let the given annulus be $r_{2}<\left|{z-z_{0}}\right|< r_{1}$ where $r_{1}$ and $r_{2}$ are the radii of $c_{1}$ and $c_{2}$ respectively (Fig. 370). This corresponds to ${{1}\over{r_{2}}}>\left|{Z}\right|>{{1}\over{r_{1}}}$ Hence this power series in Z must converge at least in the disk $\left|{Z}\right|<{{1}\over{r_{2}}}$ This corresponds to the exterior $\left|{z-z_{0}}\right|> r_{2}$ of $c_{2}$ so that $h(z)$ is analytic for all z outside $c_{2}$. Also, $h(z)$ must be singular inside $c_{2}$ where $f(z)$ is singular, and the series of the negative powers converges for all z in the exterior $E$ of the circle with center $z_{0}$ and radius equal to the maximum distance from $z_{0}$ to the singularities of $f(z)$ inside $c_{2}$. The domain common to $D$ and $E$ is the enlarged open annulus characterized near the end of Laurent’s theorem, whose proof is now complete.
Now, the problem is this: since the function $g(z)$ is equal to the series $$\sum\limits_{n=0}^{\infty}{a_{n}(z-z_{0})^{n}}$$ which is convergent in disk D so g(z) is analytic in D. Also h(z) is equal to the series $$\sum\limits_{n=1}^{\infty}{{{b_{n}}\over{(z-z_{0})^{n}}}}$$ which converges in the exterior $E$ of the circle with center $z_{0}$ and radius equal to the maximum distance from $z_{0}$ to the singularities of $f(z)$ inside $c_{2}$. so h(z) is analytic in E. from this discussion, we conclude that the singularities of f(z) outside $c_{1}$ must be caused by g(z) and singularities of f(z) inside $c_{2}$ must be caused by h(z) as the book said before. the book says: $g(z)$ must be singular at all points outside $c_{1}$ where $f(z)$ is singular. but we defined g(z) inside $c_{1}$ and we did not define it outside $c_{1}$ in other words the series which defines g(z) is convergent only inside D so how it can be defined outside $c_{1}$? in the same way h(z) is defined in the exterior E so how the book says:$h(z)$ must be singular inside $c_{2}$ where $f(z)$ is singular? [the textbook is ''Advanced Engineering Mathematics by Erwin Kreyszig'' 10th edition pages 709-712]
You can maybe think about it like this:
We use Taylor series and Laurent series to "approximate" functions, (if we use the whole series than it is not an approximation of course).
And $g(z)$ is the Taylor series that converges from the point from where we develop the "approximation" to the first singularity. On that distance from the starting point the "approximation" brakes apart. And on that circle it diverges.
After that point of singularity you use $h(z)$ which is the Laurent series and is converges from that point "forward".
So Taylor series converges from the "starting point" up to the first singularity, and Laurent series converges from the singularity and forward.
So the red is the Taylor series approximation and green is the Laurent series approximation.
(Once again if we use the whole series than it is not an approximation but an equality).
I don't know if this will help, but maybe you can visualise it like this.