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}
