As the question title asks for, how do others visualize the RiemannRoch theorem with complex analysis or geometric topology considerations? That is all Riemann would have had back in the day, and he reminds me of Thurston in the sense he just drew a picture and then the proof popped out. Nowadays most formulations of RiemannRoch are couched in algebraic language and do not invoke any geometric intuition for the complex plane, so I am asking here. Bonus points for pictures.

3$\begingroup$ "he reminds me of Thurston in the sense he just drew a picture and then the proof popped out"  citation needed $\endgroup$ – Yemon Choi Feb 27 at 0:22

$\begingroup$ @YemonChoi I have heard Dennis Sullivan and his acolytes say this before. $\endgroup$ – Get Off The Internet Mar 5 at 18:57

$\begingroup$ Were they talking about Riemann, or about Thurston? $\endgroup$ – Yemon Choi Mar 5 at 19:44

1$\begingroup$ Are you specifically asking for how Riemann might have thought geometrically about his proof of a special case of what is now known as the RR theorem, or a more general question about possible geometric or complexanalytic intuition for the general, modern version(s) of the RR theorem? $\endgroup$ – Yemon Choi Mar 5 at 19:51

$\begingroup$ If you're asking about what Riemann did, then Roy Smith wrote a nice answer here mathoverflow.net/questions/253090/…  my apologies if you know all this and are explicitly asking about more general versions $\endgroup$ – Yemon Choi Mar 5 at 19:52
I'll try to give this a shot, but I'm coming from an algebraic geometry perspective. In the case of curves, it's not so clear to me what the line between algebraic geometry and complex geometry is, so I'm not sure if what I'm saying is what you're looking for. Anyways, let's assume that our curve $C$ is not hyperelliptic because I want to use the canonical embedding because embeddings are easier to visualize for me. The RiemannRoch theorem says for a divisor $D$ on $C$ that $$ h^0(D)=1g+\deg(D)+h^0(KD) $$ We know that $H^0(KD)$ is measuring sections of the canonical divisor $K$ that vanish on $D$. But in the canonical embedding, $H^0(C,K)=H^0(\mathbb{P}^{g1},\mathcal{O}(1))$. Hence, sections of the canonical divisor are just linear homogeneous polynomials on a projective space. Therefore, $H^0(KD)$ is just hyperplanes in the canonical embedding that contain $D$. Now consider the linear span of $D$, which is some projective space $\mathbb{P}^k\cong \text{Span}(D)$. The linear polynomials that vanish on $D$ are the same as the linear polynomials vanishing on its span, $\mathbb{P}^{k}$, and we can compute the number of these. There are $g1k$ of them. Plugging back into RiemannRoch, we get $$ h^0(D)=\deg D\dim \text{Span}(D) $$ I think this statement is more amenable to visualizing because the right hand side just asks you to place a bunch of points on a curve in projective space, and then take their linear span. I have no idea if this is how Riemann thought about it, but it's certainly geometric. If you want to read more, I learned about this from Geometry of Algebraic Curves Volume 1 by Arbarello, Cornalba, Griffiths, and Harris.
Complex geometric view of Riemann Roch for a curve $C$:
The essential Riemann Roch problem is the computation of the dimension of the vector space $H^0(D)$ where $D$ is an effective divisor on $C$. The first and most fundamental case is that for the canonical divisor $K$, whose dimension is $\dim H^0(K) = g = \operatorname{genus}(C)$. Geometrically this says that the Jacobian variety $J(C) = H^0(K)^*/H_1(C; \mathbb{Z})$ of the curve is a compact complex torus of dimension $g$. Then the geometry of the RR problem for the general case of an effective divisor $D$ of degree $d$, is encoded by the AbelJacobi map $C^{(d)} \to J(C)$, from the $d$fold symmetric product of the curve to the Jacobian. This is based on the theorem of Abel, which says the fiber of this map containing $D$, is exactly the projective space built on the vector space $H^0(D)$, i.e. the fiber is the linear series $D$. Thus $\dim H^0(D) = 1 + \dim D = 1 + $ dimension of the fiber containing $D$.
Since the Abel map is holomorphic, this implies that $\dim D \geq \dim C^{(d)}  \dim J(C) = dg$. Thus $\dim H^0(D)  1 \geq d g$, or $\dim H^0(D) \geq 1g+d$, which is the “weak” Riemann Roch theorem, i.e. the version usually attributed to Riemann.
Then Roch’s augmentation computes the rank of the derivative of the Abel map. I.e. using the fact that by definition the canonical map on the curve, mentioned in Samir Canning's answer, is essentially the derivative of the Abel map, it follows that the linear rank of the Abel map, i.e. the rank of its derivative at $D$, equals (one more than) the dimension of the span of the points of $D$ in canonical space, i.e. $g  \dim H^0(KD)$. Roch’s result is that the derivative of the Abel map collapses only tangent vectors along the fiber, so that this rank also equals the non linear rank, i.e. $d  \dim D$. Thus we have $d\dim D = g\dim H^0(KD)$, hence $\dim H^0(D) = 1 + \dim D = 1g+d + \dim H^0(KD)$, which is the full RiemannRoch theorem.
Thus looking at the Abel map $C^{(d)} \to J(C)$, the RR problem is the computation of the dimension of the fibers, and the RRT says that the dimension of the fiber of the Abel map through $D$, equals the dimension of the fiber i.e. of the kernel, of the derivative of the Abel map at $D$. This equality (after adding one to both sides) is expressed in the second formula in Samir's answer, i.e. the geometric RRT. I.e. Samir's discussion essentially explains why the right hand side of his formula equals (one plus) the nullity of the derivative of the Abel map, and I have tried to motivate why it should equal the left hand side, i.e. (one plus) the dimension of D. Put another way, RRT just says the fiber of the Abel map through D is isomorphic to D as (non singular) schemes, which is just a slight strengthening of Abel's theorem.
This point of view is explained in the classic paper of Mattuck and Mayer, where it is used to give a proof of RRT, modulo the existence of the Jacobian variety. https://eudml.org/doc/83304
Here is my "visual" coming from the complex (and functional) analysis perspective, due to a novel proof by Taubes. For a holomorphic line bundle $E$ over a genus $g$ surface $C$, the RR theorem states that $ind_\mathbb{C}(\bar\partial)=c_1(E)+1g$.
Because the index of a Fredholm operator over a closed manifold does not change if we perturb by a 0th order (compact) operator, $ind_\mathbb{R}(\bar\partial)=ind_\mathbb{R}(\bar\partial+rB)$ for every $r\in\mathbb{R}_+$ and generic section $B\in\Gamma(E^2\otimes T^{0,1}C)$. Here $B$ acts as a complex antilinear map in order to define the perturbation of $\bar\partial:\Gamma(E)\to\Gamma(E\otimes T^{0,1}C)$.
Now we can visualize the (co)kernel of our perturbed operator as $r\to\infty$, because the elements of the (co)kernel localize/concentrate around the zeros of $B$. Thus $ind_\mathbb{R}(\bar\partial)=\# B^{1}(0)$.
Now we can visualize and appeal to the "geometry" of characteristic classes and sections of bundles: $\# B^{1}(0)=c_1(E^2\otimes T^{0,1}C)$, which equals $2c_1(E)+\chi(C)$.

$\begingroup$ I'm too lazy to draw pictures, but it's possible! $\endgroup$ – Chris Gerig Feb 26 at 19:48

$\begingroup$ Very interesting approach! To which paper by Taubes are you referring to exactly? $\endgroup$ – Warlock of Firetop Mountain Mar 5 at 20:13