Here is how I think about G-R-R in the context of moduli of curves. I realize now that I wrote something quite long.
Let me recall first the definition of the tautological ring. As a consequence of the results on the birational geometry of $\overline M_g$ that there is no hope of understanding the whole Chow ring of $\overline M_g$ -- for instance, unlike in misleading low genus examples, the Chow ring will in general be infinite-dimensional. In David Mumford's "Towards an enumerative geometry..." he he introduces a finite-dimensional subring of the Chow ring which contains all "geometrically natural" classes in the Chow ring and proposes studying it instead, and this subring is called the tautological ring. Let me quote:
"Whenever a variety or topological space is defined by some universal property, one expects that by virtue of its defining property, it possesses certain cohomology classes called tautological classes. The standard example is a Grassmannian [...] by its very definition, there is a universal bundle $E$ on Grass of rank $k$, and this induces Chern classes $c_l(E)$ in both the cohomology ring and Chow ring of Grass."
Let me expand on the meaning of "geometrically natural". There are several possible definitions of the tautological ring. The one used by Mumford is that it is the subring generated by the so-called $\kappa$-classes, which is not really the right one: you should also for instance consider the boundary divisors as tautological classes (but this is implicit already in Mumford's paper). A nice definition is the one of Faber and Pandharipande, which defines the tautological ring for all spaces $\overline M_{g,n}$ simultaneously: it is the minimal system of subrings which contains all fundamental classes, is closed under all gluing morphisms, and is closed under all forgetting points-morphisms.
Morally what this means is that: (i) for any "natural" bundle you can write down directly in terms of the moduli functor, its Chern classes are going to be tautological; (ii) any sort of "natural" gluing procedure on curves is going to keep you inside of the tautological ring. For example, the $\lambda$-classes (Chern classes of the Hodge bundle) are tautological, the $\psi$-classes are tautological (the line bundles given by the cotangent line at a marked point), and the $\kappa$-classes are tautological.
OK, so let us return to G-R-R. Let $f \colon X \to Y$ be a proper morphism. On one side of the equation you have the Chern character of the derived pushforward $Rf_\ast F$. On the other side you have the pushforward of the Chern character of $F$ and the Todd class of the relative tangent sheaf $T_f$. The point is that both $F$, $Rf_\ast F$ and $T_f$ can all be made sense of by working locally/fiberwise: we don't need to know anything about the global structure of $Y$ to apply G-R-R to $f$ and $F$. But this is also how the tautological ring was set up: the classes in the tautological ring are exactly those that can be defined by pushing around classes of "fiberwise" defined bundles, which means that these are exactly the classes that can be defined without making any reference to any "global" structure of the moduli space.
So in hindsight Grothendieck-Riemann-Roch seems tailor made for the study of tautological rings. On the other hand, this is also a limitation of G-R-R: it will produce lots of relations and identities relating tautological classes to each other, but it will never prove any "global" statement about any of them.
As an example, it is possible to algorithmically compute the intersection number on $\overline M_{g,n}$ for any polynomial in boundary strata and $\lambda$-, $\psi$- and $\kappa$-classes. First you express the $\kappa$-classes as pushforwards of $\psi$-classes, then G-R-R can be used to express the $\lambda$-classes in terms of pushforwards of $\psi$-classes, which will finally reduce your computation to an intersection number only involving $\psi$-classes. All this was completely formal, but sooner or later you are going to need to use some global geometric property of $\overline M_{g,n}$ to find an actual number, and this is where it comes in: the Witten conjecture/Kontsevich's theorem tells you how to compute any intersection of $\psi$-classes.
So let me finally talk a bit about the article of Harris and Mumford. The first application of G-R-R in their article is to derive the formula $K_{\overline{M}_g} = 13\lambda_1 - 2\delta_0 - 3\delta_{1} - \ldots - 2\delta_{n}$ in the tautological ring. This is done by applying GRR to the projection from the universal curve and truncating after the first term. Incidentally, if you don't truncate after the first term, you get Mumford's formula (derived in "Towards an enumerative geometry...") expressing the Chern character of the Hodge bundle in terms of $\kappa$-classes and pushforwards of $\psi$-classes from the boundary strata.
But again, GRR will not tell you any global geometric information like if a class is big or ample. The idea is then to find an effective divisor $D$ such that $mK_{\overline M_g} = D + a\lambda_1$ with $a > 0$. It turns out that this is possible for $D$ equal to the locus of $k$-gonal curves, where they pick $g = 2k-1$. They describe in the article how they came up with this particular choice of $D$ by trying to generalize the work of Freitag on the Kodaira dimension of $A_g$ for $g$ large, in particular I think that there should be a Siegel modular form whose pullback to $M_g$ conjecturally would have $D$ as its vanishing locus. I don't know if this was actually worked out in later work. Then $nK_{\overline M_g}$ for large enough $n$ defines a birational map using the fact that $\lambda_1$ is ample on $A_g$, ultimately because the Satake compactification is the Proj of the ring of Siegel modular form, i.e. the sections of powers of the determinant of the Hodge bundle. (However $\lambda_1$ is not ample on $\overline M_g$!)
This part is clarified by the later article of Cornalba and Harris showing that a linear combination $a\lambda - b\delta$ is ample if and only if $a > 11b$. The rational Picard group of $\overline M_g$ is generated by $\lambda_1$ and the boundary divisors, so any effective divisor has an expression of the form $a\lambda - \sum b_i \delta_i$, so estimating the Kodaira dimension of $\overline M_g$ really comes down to finding effective divisors such that the slopes $a/b_i$ are small.
Anyway, the second application of GRR in their article is to show that on the open part $M_g$, the $k$-gonal locus is a multiples of $\lambda_1$. Actually, this part uses even more crucially Porteous's formula: once they express $k$-gonality in terms of a morphism of bundles having lower than expected rank, the class of the $k$-gonal locus can be expressed in terms of Chern classes of the two bundles, i.e. in terms of tautological classes. It follows then that $D$ is the sum of a multiple of $\lambda_1$ and an integral linear combination of boundary divisors. Finally these integers are determined by evaluating the divisors on suitable "test curves". They conclude that $\overline M_g$ is of general type for big $g$.
Yes.
Let $C$ be a hyperelliptic curve of genus $g$, and let $L$ be a general line bundle of degree $g+1$. By Riemann-Roch, $\dim|L| = 1$ and $|L|$ is base-point free, so the complete series $|L|$ gives a degree $g+1$ map to $\mathbb{P}^1$. Then the product of this map and the degree $2$ map $C\to \mathbb{P}^1$ gives a map $f:C\to \mathbb{P}^1 \times \mathbb{P}^1$, whose image is a curve of type $(2,g+1)$. But $\mathbb{P}^1\times \mathbb{P}^1$ is just a quadric in $\mathbb{P}^3$.
To see the map is an embedding, it will suffice to show that it is birational. Indeed, the image has arithmetic genus $g$ by adjunction on a quadric surface. But it also has geometric genus $g$ since $C$ is its normalization. Thus the image is smooth if it is reduced.
Finally we must see that the map is birational. The only way the map fails to be injective is if some divisor of $|L|$ contains a pair of points conjugate under the hyperelliptic involution. But in this case the assumption that $\dim |L|=1$ implies that $|L| = g_2^1 + p_1+\cdots+ p_{g-1}$, where $g_2^1$ is the hyperelliptic series and $p_1,\ldots,p_{g-1}$ are base points. Since $L$ was general, this isn't true, and we're done.
Here's a cute (although trivial) kind of partial converse: If $g$ is prime, then any smooth curve $C$ of genus $g$ which embeds in a smooth quadric is hyperelliptic.
Best Answer
Suppose that $D=x_1+x_2+\cdots+x_{g-1}$. We may assume that $\tau(x_i)\neq x_j$ for all $i\neq j$. Now, assume that $D'=y_1+\cdots+y_{g-1}$ is an element of $|K-D|$. Then $x_1+x_2+\cdots+x_{g-1}+y_1+\cdots+y_{g-1}$ is an element of $|K|$ but we know that any such element is of the form $z_1+\tau(z_1)+\cdots+z_{g-1}+\tau(z_{g-1})$. After possibly renumbering the $z_i$ (as well as possibly replacing $z_i$ by $\tau(z_i)$) we may assume $x_1=z_1$ and then $x_2=z_2$ and so on. This gives $D'=\tau(D)$ which implies $h^0(K-D)\leq 1$. However, R-R and gives $h^0(D)=h^0(K-D)\leq1$ which contradicts assumptions.