Since you are curious about 'what is out there', let me offer this slightly different way to think about this problem: To say that $f$ is a linear fractional function of $z$ is to say that there are constants $(a,b,c,d)$ with $ad-bc\not=0$ such that  $$ c\ z f + d\ f - a\ z - b = 0. $$ In other words, one is requiring that the mapping $H_{f,z}:\mathbb{D}\to \mathbb{P}^3$ defined by $$ H_{f,z} = [1, f, z, fz] $$ should lie in a hyperplane $\mathbb{P}^2\subset\mathbb{P}^3$. The condition that $f$ not be constant is what keeps this map from going into a line in $\mathbb{P}^3$. Note, though, that the image of $H_{f,z}$ always lies in the quadric $\mathbb{Q}\subset\mathbb{P}^3$ defined by $X_0X_3-X_1X_2=0$, and that this quadric is biholomorphic to $\mathbb{P}^1\times\mathbb{P}^1$ in the obvious way: $\bigl([a_0,a_1],[b_0,b_1]\bigr)\mapsto [a_0b_0,a_1b_0,a_0b_1,a_1b_1]$.

This suggests the following generalization: Let $D$ be any connected Riemann surface and let $f$ and $g$ be nonconstant meromorphic functions on $D$. Define $H_{f,g}:D\to \mathbb{Q}\subset\mathbb{P}^3$ by $$ H_{f,g} = [1, f, g, fg]. $$ Then this curve does not have image in a line (since $f$ and $g$ are not constant). The condition that $H_{f,g}$ have image lying in a plane is that the relative Schwarzian $\mathsf{S}(f,g)$ should vanish, where $$ \mathsf{S}(f,g) = \left( \frac{f'''}{f'}-\frac32\frac{(f'')^2}{(f')^2} -\frac{g'''}{g'}+\frac32\frac{(g'')^2}{(g')^2}\right)\ dw^2 $$ where the primes denote differentiation with respect to any local coordinate $w$. (It is easy to verify that the meromorphic quadratic differential $\mathsf{S}(f,g)=-\mathsf{S}(g,f)$ is well-defined, independent of the choice of local coordinate $w$.)

Now, if you want a quantitative sense of how close $H_{f,g}$ is to lying in a plane, you are asking for a quantitative sense of how close the osculating map $$ K_{f,g} = [H_{f,g}\wedge H'_{f,g}\wedge H''_{f,g}]:D\to \mathbb{P}^3 $$ is to being constant. (Here, I am using the standard fact that $\mathbb{P}^3 = \mathbb{P}(\mathbb{C}^4)$; that, up to a scalar multiple, one has $\Lambda^3(\mathbb{C}^4)\simeq (\mathbb{C}^4)^\ast$; and that, because there is a nondegenerate quadratic form specified on the $\mathbb{C}^4$, we can identify $(\mathbb{C}^4)^\ast$ with $\mathbb{C}^4$. Thus, one can interpret the trivector in the brackets as an ordinary vector, up to multiples, and this multiple goes away when you projectivize. Also, the choice of local parameter $w$ used to compute the derivatives clearly will not matter, so that $K_{f,g}$ is well-defined.)

To get a sense of how close $K_{f,g}$ is to being constant, fix some metric on $\mathbb{P}^3$; for the purposes of this argument, say we fix the Fubini-Study metric with Kähler form $\Omega$ on $\mathbb{P}^3$. (This is where we are introducing symmetry breaking, since there is no canonical metric on $\mathbb{P}^3$, but rather a family of Fubini-Study metrics, all differing by projective transformations, so the moduli space of such is $\mathrm{SL}(4,\mathbb{C})/\mathrm{SU}(4)$. You can do a little better than this because you can choose one that is invariant under the maximal compact in the automorphism group of the inner product, and that moduli space is $\mathrm{SO}(4,\mathbb{C})/\mathrm{SO}(4)$, and that is what I would recommend.)

At any rate, consider the nonnegative real $(1,1)$-form on $D$ defined by  $$ \Omega_{f,g} = (K_{f,g})^*(\Omega). $$ This $(1,1)$-form vanishes identically if and only if $H_{f,g}$ has image lying in a plane, otherwise, it vanishes only at isolated points. Indeed, when it doesn't vanish identically, relative to a local coordinate $w$ on $D$, it takes the form $$ \Omega_{f,g} = \frac{i}{2}|w|^{2k}p(w,\bar w)\ dw\wedge d\bar w. $$ where $p$ is nonvanishing and $k\ge0$. In particular, the (possibly degenerate) metric given in local coordinates by $$ ds^2_{f,g} = |w|^{2k}p(w,\bar w)\ dw\circ d\bar w $$ is well-defined on $D$. Now, in fact, $ds^2_{f,g}$ is essentially the square norm of $\mathsf{S}(f,g)$ times a factor that 'clears the denominators', so that $ds^2_{f,g}$ does not have 'poles'. Thus, in a natural sense, a bound on $ds^2_{f,g}$ is equivalent to a bound on $\mathsf{S}(f,g)$. (Bear in mind that this is not really quite true because I have broken symmetry by choosing $\Omega$.)

In particular, if you were to take the 'norm' on $\mathsf{S}(f,g)$ to be the diameter of $D$ with respect to the singular metric $ds^2_{f,g}$ (which is well-defined), then that norm would give you an upper bound on the diameter of the image of $K_{f,g}$, i.e., it would say that, for example, the distance from $K_{f,g}(p)$ to $K_{f,g}(p_0)$ is bounded, so that the image of $H_{f,g}$ would lie at most a bounded distance from the plane that $K_{f,g}(p_0)$ defines.

  In geometric terms, this is exactly the kind of bound you were asking for. The only difference is that you aren't using a norm directly on the Schwarzian, but on a closely related quantity $ds^2_{f,g}$, one that depends on an arbitrary choice (so it's not quite canonical) but that has better compactness properties (so that you can actually get estimates).

Here is a revised and somewhat expanded version of my answer, with a preparatory 'toy version' to help orient the reader.

A simple warmup problem: Before discussing a quantitative variant of the Schwarzian, let me describe the overall idea in a simpler case: Deciding how close two nonconstant meromorphic functions on a Riemann surface are to being constant multiples of each other.

Let $D$ be a connected Riemann surface, and let $f$ and $g$ be nonconstant meromorphic functions on $D$. They will be constant multiples of each other if the meromorphic $1$-form $\omega = f\ dg - g\ df$ vanishes identically. However, this is an exact statement, and it is hard to see how a bound on the 'size' of $\omega$ (assuming that it doesn't vanish outright) could say anything quantitative about how 'close' $f$ and $g$ are to being constant multiples of each other, especially since $f$ and $g$ could have poles (and $\omega$ could have poles, too, for that matter).

As an alternative, consider the singular (i.e., possibly degenerate) conformal metric on $D$ defined by $$ ds^2_{f,g} = \frac{4\ (f\ dg - g\ df)\circ\overline{(f\ dg - g\ df)}} {\bigl(|f|^2+|g|^2\bigr)^2}\ . $$ This is the singular metric induced on $D$ by pullback under the holomorphic map $H_{f,g}:D\to\mathbb{CP}^1$ defined by $H_{f,g} = [f,g]$, where the metric on $\mathbb{CP}^1$ is the familiar Fubini-Study metric with constant Gauss curvature 1: $$ d\sigma^2 = \frac{4\ (z\ dw - w\ dz)\circ\overline{(z\ dw - w\ dz)}} {\bigl(|z|^2+|w|^2\bigr)^2}\ . $$

Note that $ds^2_{f,g}$ vanishes identically when $\omega$ does but never has poles, so it's easier to define and measure its 'size'. Most importantly, a bound on the size of $ds^2_{f,g}$ implies a bound on the size of the image of $H_{f,g}$. For example, if $R>0$ is the diameter of $D$ under the metric $ds^2_{f,g}$, then the image of $H_{f,g}$ fits into a disk of radius at most $R$ in $\mathbb{CP}^1$. Obviously, the smaller $R$ is, the closer $H_{f,g}$ is to being a constant map, i.e., the closer the ratio of $f$ to $g$ is to being constant. Meanwhile, there are many ways to estimate the diameter of $D$ under the metric $ds^2_{f,g}$; for example, one could have a pointwise bound of the ratio of this metric to another 'reference' metric whose diameter is already known.

Now, the noncanonical thing about this strategy is that there is more than one candidate for the Fubini-Study metric on $\mathbb{CP}^1$. However, these are all equivalent under automorphisms of $\mathbb{CP}^1$, so they form a family parametrized by $\mathrm{SL}(2,\mathbb{C})/\mathrm{SU}(2)$ (which is, of course hyperbolic $3$-space). Moreover, the ratio of any two of them is bounded, so, in a sense, switching to a different member of the family will produce a quantitative measurement that is not essentially

different.

**A Quantitative version of the Schwarzian:** With the above in mind, consider this geometric

way to think about the problem of quantifying the Schwarzian: To say that $f$ is a linear fractional function of $z$ is to say that there are constants $(a,b,c,d)$ with $ad-bc\not=0$ such that  $$ c\ z f + d\ f - a\ z - b = 0. $$ In other words, one is asking whether the mapping $H_{f,z}:\mathbb{D}\to\mathbb{P}^3$ defined by $$ H_{f,z} = [1, f, z, fz] $$ has image in a hyperplane $\mathbb{P}^2\subset\mathbb{P}^3$. The condition that $f$ not be constant is what keeps this map from going into a line in $\mathbb{P}^3$.

Meanwhile

, note that the image of $H_{f,z}$ always lies in the quadric $\mathbb{Q}\subset\mathbb{P}^3$ defined by $X_0X_3-X_1X_2=0$, and that this quadric is biholomorphic to $\mathbb{P}^1\times\mathbb{P}^1$ in the obvious way: $\bigl([a_0,a_1],[b_0,b_1]\bigr)\mapsto [a_0b_0,a_1b_0,a_0b_1,a_1b_1]$. In particular, we should be thinking of $\mathbb{P}^3$ as the projectivization of the vector space $V\simeq\mathbb{C}^4$ of $2$-by-$2$ matrices endowed with the quadratic form $X_0X_3-X_1X_2$ that is simply the determinant. For use below, I will also fix a volume form on $V$. That way, one can define a triple cross product of vectors $v_1,v_2,v_3\in V$ by letting $v_1\times v_2\times v_3\in V$ be the vector that satisfies $$ (v_1\times v_2\times v_3)\cdot w = \det(v_1\wedge v_2\wedge v_3\wedge w) $$ for all $w\in V$.

Now, let

$D$ be any connected Riemann surface and let $f$ and $g$ be nonconstant meromorphic functions on $D$. Define a holomorphic curve $H_{f,g}:D\to \mathbb{Q}\subset\mathbb{P}^3$ by $$ H_{f,g} = [1, f, g, fg]\ . $$ Then this curve does not have image in a line (since $f$ and $g$ are not constant). The condition that $H_{f,g}$ have image lying in a plane is that the *relative Schwarzian* $\mathsf{S}(f,g)$ should vanish, where $$ \mathsf{S}(f,g) = \left( \frac{f'''}{f'}-\frac32\frac{(f'')^2}{(f')^2} -\frac{g'''}{g'}+\frac32\frac{(g'')^2}{(g')^2}\right)\ dw^2 $$ and where the primes denote differentiation with respect to any local coordinate $w$. (It is easy to verify that the meromorphic quadratic differential $\mathsf{S}(f,g)=-\mathsf{S}(g,f)$ is defined independent of the choice of local coordinate $w$.)

To get

a quantitative sense of how close $H_{f,g}$ is to lying in a plane, one needs a quantitative sense of how close the 'normal' map $$ K_{f,g} = [{\hat H}_{f,g}\times {\hat H}'_{f,g}\times {\hat H}''_{f,g}]: D\to\mathbb{P}^3 $$ is to being constant. (Here, I am taking a (local if necessary) meromorphic lifting ${\hat H}_{f,g}:D\to V$. This lifting is unique up to a multiple, and this multiple goes away under projectivization. Also, the choice of local parameter $w$ used to compute the derivatives does not matter, so that $K_{f,g}$ is well-defined.)

 

To get a sense of how close $K_{f,g}$ is to being constant, fix some metric on $\mathbb{P}^3$; for the purposes of this argument, say we fix a Fubini-Study metric with Kähler form $\Omega$ on $\mathbb{P}^3$. (This is where we are introducing symmetry breaking, since there is no canonical metric on $\mathbb{P}^3$, but rather a family of Fubini-Study metrics, all differing by projective transformations, so the moduli space of such is $\mathrm{SL}(4,\mathbb{C})/\mathrm{SU}(4)$. One can do a little better than this because one can choose one that is invariant under the maximal compact in the automorphism group of the inner product, and that smaller moduli space is $\mathrm{SO}(4,\mathbb{C})/\mathrm{SO}(4)$.)

  At any rate, consider the nonnegative real $(1,1)$-form on $D$ defined by  $$ \Omega_{f,g} = (K_{f,g})^*(\Omega). $$ This $(1,1)$-form vanishes identically if and only if $H_{f,g}$ has image lying in a plane, otherwise, it vanishes only at isolated points. Indeed, when it doesn't vanish identically, relative to a local coordinate $w$ on $D$, it takes the form $$ \Omega_{f,g} = \frac{i}{2}|w|^{2k}p(w,\bar w)\ dw\wedge d\bar w. $$ where $p$ is nonvanishing and $k\ge0$. In particular, the (possibly degenerate) metric given in local coordinates by $$ ds^2_{f,g} = |w|^{2k}p(w,\bar w)\ dw\circ d\bar w $$ is well-defined on $D$. Calculation shows that $ds^2_{f,g}$ is essentially the square norm of $\mathsf{S}(f,g)$ times a factor that 'clears the denominators', so that $ds^2_{f,g}$ does not have 'poles'. Thus, in a natural (but not canonical) sense, a bound on $ds^2_{f,g}$ is equivalent to a bound on $\mathsf{S}(f,g)$. 

In particular, if you were to take the 'norm' on $\mathsf{S}(f,g)$ to be the diameter of $D$ with respect to the singular metric $ds^2_{f,g}$ (which is well-defined), then that norm would give you an upper bound on the diameter of the image of $K_{f,g}$, i.e., it would say that, for example, the distance from $K_{f,g}(p)$ to $K_{f,g}(p_0)$ is bounded, so that the image of $H_{f,g}$ would lie at most a bounded distance from the plane that $K_{f,g}(p_0)$ defines. 

In geometric terms, this is exactly the kind of bound you were asking for. The only difference is that you aren't using a norm directly on the Schwarzian, but on a closely related quantity $ds^2_{f,g}$, one that depends on an arbitrary choice (so it's not quite canonical) but that has better compactness properties (so that you can actually get estimates). 

This suggests the following generalization: Let $D$ be any connected Riemann surface and let $f$ and $g$ be nonconstant meromorphic functions on $D$. Define $H_{f,g}:D\to \mathbb{Q}\subset\mathbb{P}^3$ by $$ H_{f,g} = [1, f, g, fg]. $$ Then this curve does not have image in a line (since $f$ and $g$ are not constant). The condition that $H_{f,g}$ have image lying in a plane is that the relative Schwarzian $\mathsf{S}(f,g)$ should not vanish, where $$ \mathsf{S}(f,g) = \left( \frac{f'''}{f'}-\frac32\frac{(f'')^2}{(f')^2} -\frac{g'''}{g'}+\frac32\frac{(g'')^2}{(g')^2}\right)\ dw^2 $$ where the primes denote differentiation with respect to any local coordinate $w$. (It is easy to verify that the meromorphic quadratic differential $\mathsf{S}(f,g)=-\mathsf{S}(g,f)$ is well-defined, independent of the choice of local coordinate $w$.)

This suggests the following generalization: Let $D$ be any connected Riemann surface and let $f$ and $g$ be nonconstant meromorphic functions on $D$. Define $H_{f,g}:D\to \mathbb{Q}\subset\mathbb{P}^3$ by $$ H_{f,g} = [1, f, g, fg]. $$ Then this curve does not have image in a line (since $f$ and $g$ are not constant). The condition that $H_{f,g}$ have image lying in a plane is that the relative Schwarzian $\mathsf{S}(f,g)$ should vanish, where $$ \mathsf{S}(f,g) = \left( \frac{f'''}{f'}-\frac32\frac{(f'')^2}{(f')^2} -\frac{g'''}{g'}+\frac32\frac{(g'')^2}{(g')^2}\right)\ dw^2 $$ where the primes denote differentiation with respect to any local coordinate $w$. (It is easy to verify that the meromorphic quadratic differential $\mathsf{S}(f,g)=-\mathsf{S}(g,f)$ is well-defined, independent of the choice of local coordinate $w$.)