Here's how I think about Laurent series, which might be helpful.
If $f$ is complex-differentiable on $R_1<|z-a|, then any (positively oriented) circle $C_r$ of radius $r$ inside the annulus (i.e. such that $R_1 < r < R_2$) defines a function $f_r(z)=\frac1{2\pi i}\oint_{C_r}\frac{f(\zeta)}{\zeta-z}d\zeta$ analytic everywhere on the disk $|z-a| except the circle $C_r$ itself where the integral is undefined. Because these circles cannot be deformed to a single point it is not necessarily true that $f_r(z)$ equals $f(z)$ for points $z$ inside the circle $C_r$.
Now, if we think about a bigger circle C_{r'} of radius r', i.e. such that r, then f_{r'}(z)=f_r(z) whenever $z$ is inside the smaller circle $C_r$ since we can deform C_{r'} onto $C_r$ without passing through such a $z$, i.e. the two circles are homotopic on the annulus minus $z$, which means that the integrals $\oint_{C_r}\frac{f(\zeta)}{\zeta-z}d\zeta$ and \oint_{C_{r'}}\frac{f(\zeta)}{\zeta-z}d\zeta are equal since $\frac{f(\zeta)}{\zeta-z}$ is complex-differentiable function (of $\zeta$!) everywhere on the annulus except the specific point $z$.
Hence, the power series of $f_r(z)$ around $a$ for any $0 converges on the whole disk $|z-a| and defines an analytic function $f_1$ on that disk $|z-a|. The Taylor Series $f(z)=\sum_{i=0}^\infty a_i(z-a)^i$ in fact gives the the non-negative power part of the Laurent series for $f(z)$ on that annulus, and $f_2$, once we've defined it, will give the negative power part.
So what can we say about $f(z)-f_1(z)$, which is definitely complex-differentiable on that annulus? Well, we can say that $\oint_{C_r}\frac{f(\zeta)-f_1(\zeta)}{\zeta-z}d\zeta=\oint_{C_r}\frac{f(\zeta)}{\zeta-z}-\frac{f_1(\zeta)}{\zeta-z}d\zeta=f_1(z)-f_1(z)=0$ for every $z$ inside $C_r$, which means we cannot use the same trick as before.
But wait! We have forgotten half of the informations given to us by the $f_r(z)$ since $f_1(z)=f_r(z)$ only when $|z-a| -- what happens when $|z-a|>r$ (recall that $f_r(z)$ is analytic everywhere except the circle $C_r$ given by $|z-a|=r$)? Well, just as before the deformations tell us that f_r(z)=f_{r'}(z) for |z-a|>r'>r so the $f_r(z)$ also define an analytic function $f_2$ on the disk-complement $|z-a|>R_1$ (while $f_1$ was defined on |z-a| Note that in particular for
|z-a|>r$ we have $|f_2(z)|<\frac1{2\pi i} \oint_{C_r}\frac{|f(\zeta)|}{|\zeta-z|}=\frac 1{2\pi i}\frac MR$ where $M$ is the largest value $f$ takes on $C_r$ and $R$ is the smallest distance between $z$ and a point on $C_r$. Hence $\lim_{z\to\infty}f_2(z)=0$, so in some sense $f_2$ is in fact analytic at infinity. The sense in which $f_2$ is analytic at infinity is this. The fractional linear transformation $\phi(z)=\frac1{z-a}+a$ is complex-differentiable on the whole complex plane except at $a$, and what it does to a point $a+re^{i\theta}$ is that it sends it to $a+\frac1r e^{i-\theta}$, thus it sends the function $f_2$ analytic outside $C_{R_1}$ to a function $g$ analytic on the inside of $C_{\frac1{R_1}}$, where $g$ is given by $g(z)=f_2(\phi(z))$ for $z\neq a$. Then if the power series expansion of $f_2(\phi(z))=f(\frac1{z-a}+a)$ at $z=a$ is $\sum_{j=1}^\infty b_j(z-a)^j$, we have that in fact $f_2(z)=\sum_{j=1}^\infty b_j(z-a)^{-j}.
Hence, we can now see that f_1$ and $f_2$ are simply the unique functions analytic respectively inside the disk $|z-a|$ and outside (including $\infty$) the disk $|z-a|>R_1$ determined by the values of $f$ on the annulus, with power expansions $f_1(z)=\sum_{i=0}^\infty a_i(z-a)^i$ and $f_2(z)=\sum_{j=1}^\infty b_j(z-a)^{-j}$ respectively the non-negative and negative power part of the Laurent expansion of $f$ in that annulus.