Writing a given differential operator in terms of another given differential operator with kernel a subset of the first 
Suppose I have two differential operators $S, T$ with the property that $\ker S \subseteq \ker T$. Is it possible to exploit this property to write $T$ in terms of $S$ in some compact way?

I would be happy with an answer just for this specific example: Let $S$ be the Schwartzian derivative, $$S[f] := \frac{f'''}{f'} - \frac{3}{2}\left(\frac{f''}{f'}\right)^2,$$
and $T$ the operator
$$T[f] := \frac{f''''}{f''} - \frac{4}{3} \left(\frac{f'''}{f''}\right)^2 .$$
The kernel of $T$ consists of the rational functions of the form
$$x \mapsto \frac{a_2 x^2 + a_1 x + a_0}{b_1 x + b_0}$$
and the kernel of $S$ famously consists of the fractional linear transformations, i.e., the rational functions of the above form with $a_2 = 0$.

How can one write $T[f]$ in terms of $S[f]$ in a way that exploits the containment $\ker S \subset \ker T$?

 A: I am not sure if this qualifies as a counter-example, but let me offer it anyway.
I would say that the answer is no, there is no convenient way to do that.
Consider these differential operators on smooth functions $\mathbb R\to\mathbb R$:
$$
Af=f''\\
Bf=(1+(f')^2)f''\\
Cf%=\frac{d}{dx}\left[(1+(f')^2)f''\right]
=(1+(f')^2)f'''+2f'(f'')^2.
$$
Now $\ker A=\ker B=\{x\mapsto a+bx;a,b\in\mathbb R\}$, but you cannot express $Af(x)$ in terms of $Bf(x)$ or vice versa with any local formula.
They differ by something depending on $f'$, but there seems to be no way to write the first derivative in terms of $Af$ or $Bf$.
The operator $Cf$ can be written in terms of $Bf$ since $Cf=(Bf)'$.
But it seems that $Cf$ cannot be written in terms of $Af$, since you would also need the first derivative.
Note that $\ker B=\ker A\subsetneq\ker C$.
If you meant writing $Tf$ in terms of $Sf$ (and its derivatives) and derivatives of $f$ of order less than the degree of $Tf$, that might be possible — I assumed you meant using only $Sf$ and its derivatives.
It certainly works for my example operators and also yours: It seems to be possible to write $Tf$ using $(Sf)'$, $f'$, and $f''$.
