Does any uncertainty relation exist for interfering measurements (ie. for 2 subsequent measurements of 2 incompatible observables on the same system)? To test the usual Robertson-Schrodinger uncertainty bound, we follow the methods delineated in the answer here by Timaeus. This method talks about non-interfering measurements. For two incompatible (or compatible) observables, A and B, we do not measure both on the same system.
However, if it turns out that for some experiments, it is necessary to make two subsequent measurements of two non-commuting operators, A and B, on the same system, then the Robertson-Schrodinger uncertainty relation cannot be used anymore. For those kinds of experiments, does there exist a separate, rigorous uncertainty relation like the Robertson-Schrodinger uncertainty relation?
 A: I don't have a prior familiarity with this problem, but I found it quite interesting so I tried to scrabble around a bit and it looks like one can get a marginally non-ugly relation for the uncertainties of the kind that OP is interested in.

Let's consider the density matrix $\tilde{\rho}$ of the system after the measurement of the first operator $A$. It is easy to see that
\begin{align}
\tilde\rho&=\sum_a \mathbb{P}_a\rho\mathbb{P}_a
\end{align}
where $\rho=\vert\psi\rangle\langle\psi\vert$ is the initial (pure state) density matrix of the system, and $\mathbb{P}_a$ are the projection operators corresponding to the eigensubspaces of the operator $A$.
The measurement of the operator $A$, performed over $\rho$, has the well-known uncertainty given by
\begin{align}
\sigma_A^2 =\mathrm{Tr}\big[\rho\big(A-\langle A\rangle\big)^2\big]
\end{align}
For a more detailed calculation leading to the above expression, starting from the familiar state-vector formulation, see this answer of mine which deals with the usual RS uncertainty principle in the density matrix formulation. In order to keep the following discussion self-contained, I will freely copy relevant parts of the linked answer.
Similarly, the measurement of the operator $B$, performed over $\tilde{\rho}$ would carry an uncertainty given by
\begin{align}
\sigma_B^2 &=\mathrm{Tr}\big[\tilde\rho\big(B-\langle B\rangle\big)^2\big]\\
\end{align}
Notice that we are adopting a compact notation for the expectation value which can lead to confusion if one loses sight of the fact that $\langle B\rangle = \mathrm{Tr}(\tilde{\rho}B)$ whreas $\langle A\rangle = \mathrm{Tr}(\rho A)$.
Thus, we can write
\begin{align}
\sigma_A^2\sigma_B^2&=\mathrm{Tr}\big[\rho\big(A-\langle A\rangle\big)^2\big]\mathrm{Tr}\big[\tilde\rho\big(B-\langle B\rangle\big)^2\big]\\
\mathrm{(linearity\ of\ trace)}&=\sum_a\mathrm{Tr}\big[\rho\big(A-\langle A\rangle\big)^2\big]\mathrm{Tr}\big[\mathbb{P}_a\rho\mathbb{P}_a\big(B-\langle B\rangle\big)^2\big]\\
\mathrm{(cyclic\ property\ of\ trace)}&=\sum_a\mathrm{Tr}\big[\rho\big(A-\langle A\rangle\big)^2\big]\mathrm{Tr}\big[\rho\mathbb{P}_a\big(B-\langle B\rangle\big)^2\mathbb{P}_a\big]\\
\mathrm{(idempotent\ nature\ of\ }\mathbb{P}_a\mathrm{)}&=\sum_a\mathrm{Tr}\big[\rho\big(A-\langle A\rangle\big)^2\big]\mathrm{Tr}\big[\rho\mathbb{P}^2_a\big(B-\langle B\rangle\big)^2\mathbb{P}^2_a\big]\\
\end{align}
Now, the Cauchy-Schwarz inequalities hold true for all inner products, and it can be shown that for Hermitian operators $X$ and $Y$, it takes the following form
\begin{align}
\big|\text{Tr}\big(SX Y\big)\big|^2 \leq \text{Tr}\big(SX^2\big)\text{Tr}\big(SY^2\big)
\end{align}
where $S$ is a semi-positive definite matrix. Since $\rho$ is a semi-positive definite matrix and $A-\langle A\rangle$, and $\mathbb{P}_a(B-\langle B\rangle)\mathbb{P}_a$ are Hermitian operators, an application of this inequality yields
\begin{align}
\mathrm{Tr}\big[\rho\big(A-\langle A\rangle\big)^2\big]\mathrm{Tr}\big[\rho\mathbb{P}^2_a\big(B-\langle B\rangle\big)^2\mathbb{P}^2_a\big] &\geq \Big\vert\ \mathrm{Tr}\big[\rho\big(A-\langle A\rangle\big)\mathbb{P}_a\big(B-\langle B\rangle\big)\mathbb{P}_a\big]\ \Big\vert^2\\
\implies \sum_a\mathrm{Tr}\big[\rho\big(A-\langle A\rangle\big)^2\big]\mathrm{Tr}\big[\rho\mathbb{P}^2_a\big(B-\langle B\rangle\big)^2\mathbb{P}^2_a\big] &\geq \sum_a\Big\vert\ \mathrm{Tr}\big[\rho\big(A-\langle A\rangle\big)\mathbb{P}_a\big(B-\langle B\rangle\big)\mathbb{P}_a\big]\ \Big\vert^2 \\
\implies \sigma_A^2\sigma_B^2&\geq \sum_a\Big\vert\ \mathrm{Tr}\big[\rho\big(A-\langle A\rangle\big)\mathbb{P}_a\big(B-\langle B\rangle\big)\mathbb{P}_a\big]\ \Big\vert^2\\
\end{align}
We take a pause from our mainline calculation and manipulate the terms appearing on the RHS in the expression above. Using the benefit of the hindsight, we write
\begin{align}
\zeta_a&\equiv\mathrm{Tr}\big[\rho\big(A-\langle A\rangle\big)\mathbb{P}_a\big(B-\langle B\rangle\big)\mathbb{P}_a\big] \\
&= \mathrm{Tr}\big[\rho\big(A\mathbb{P}_aB\mathbb{P}_a-A\mathbb{P}_a\langle B \rangle \mathbb{P}_a - \langle A\rangle\mathbb{P}_aB\mathbb{P}_a+\langle A\rangle \mathbb{P}_a\langle B\rangle\mathbb{P}_a\big)\big]\\
&=\mathrm{Tr}\big[\mathbb{P}_a\rho\mathbb{P}_a\big(AB-\langle A\rangle B- \langle B\rangle A + \langle A\rangle \langle B\rangle \big)\big]\\
\implies \zeta_a^* &= \mathrm{Tr}\big[\mathbb{P}_a\rho\mathbb{P}_a\big(BA-\langle A\rangle B - \langle B\rangle A + \langle A\rangle \langle B\rangle \big)\big]\\
\implies \mathrm{Im}(\zeta_a) &= \frac{1}{2i}\mathrm{Tr}\big(\mathbb{P}_a\rho\mathbb{P}_a[A,B]\big) 
\end{align}
where we used the linearity of the trace, the cyclic property of the trace, and ${\mathrm{Tr}(A)^*}=\mathrm{Tr}(A^\dagger)$ along with the Hermiticity of the operators and the fact that $\mathbb{P}_a$ commutes with $A$ whenever necessary.
We now invoke the argument that modulus square of a complex number should be bigger than or equal to the square of its imaginary part. We can thus write
\begin{align}
\sigma_A^2\sigma_B^2&\geq \sum_a \Big\vert\ \mathrm{Im}(\zeta_a)\ \Big\vert^2 \\
\implies \sigma_A^2\sigma_B^2&\geq\frac{1}{4}\sum_a \Big\vert\ \mathrm{Tr}\big(\mathbb{P}_a\rho\mathbb{P}_a[A,B]\big)\ \Big\vert^2
\end{align}
Or, in other words,
$$\boxed{\sigma_A\sigma_B\geq\frac{1}{2}\sqrt{\sum_a \Big\vert\ \mathrm{Tr}\big(\mathbb{P}_a\rho\mathbb{P}_a[A,B]\big)\ \Big\vert^2}}$$
Notice that the traditional RS uncertainty relation in the form of density matrix reads
\begin{align}
\sigma_A\sigma_B&\geq\frac{1}{2} \Big\vert\ \mathrm{Tr}\big(\rho[A,B]\big)\ \Big\vert
\end{align}
