I've edited, just skip the first attempt and go to the second one.
THE FRAMEWORK: let us consider a real topological vector space $V$.
I thought that maybe there is a way to modify the $\widetilde{\pi}$ and $\widetilde{\sigma}$ and/or exploiting the absolute continuity of $M$ in oder to compare the LHS and the second term of RHS, and maybe writing something like $$ \left|M_{ts}^{\widetilde{\sigma_{m_{2\beta+1}}}}-M_{ts}^{\widetilde{\pi_{m_{2(\beta+1)}}}}\right| =a_{\beta}|M_{ts}^{\pi_{m_{2\beta}}}-M_{ts}^{\sigma_{m_{2\beta+1}}}| $$ for some $0<a_{\beta}<1$ such that $\limsup_{\beta\to+\infty}a_{\beta}<1$, thing that allow me to conclude, but for the moment I'm stuck.
ATTEMPT 2:
Step 1: Let us fix $m_0\ge1$ and consider $\Pi_{m_0}$ (which has $k_{m_0}+2$ elements); then, since the meshes of both sequences of partitions go to $0$, there exists $m_1>m_0$, such that, looking at $\mathfrak S_{m_1}$ we can find in it $k_{m_0}+3$ elements we label as $\{u_{j_s}^{m_1}\}_{s=0}^{k_{m_0}+2}=:\widetilde{\mathfrak S}_{m_1}$ such that \begin{align*} \left|B_{r_{l+1}^{m_0}r_l^{m_0}}-B_{u_{j_{l+1}}^{m_1}u_{j_{l}}^{m_1}}\right|&\le\frac1{2(k_{m_0}+1)^3} \;\;,\;\;\;\;l=0,\dots,k_{m_0}\\ \left|B_{u_{j_{k_{m_0}+2}}^{m_1}u_{j_{k_{m_0}+1}}^{m_1}}\right|&\le\frac1{2(k_{m_0}+1)^3}\;. \end{align*} Thus we associate step $1$ to the pair $\left(\Pi_{m_0},\widetilde{\mathfrak S}_{m_1}\right)$.
Step 2: Let us consider $\mathfrak S_{m_1}$ (which has $h_{m_1}+2$ elements); then there exists $m_2>m_1$ such that looking at $\Pi_{m_2}$, we can find in it $h_{m_1}+3$ elements we label as $\{r_{j_s}^{m_2}\}_{j=0}^{h_{m_1}+3}=:\widetilde{\Pi}_{m_2}$ such that \begin{align*} \left|B_{u_{l+1}^{m_1}u_l^{m_1}}-B_{r_{j_{l+1}}^{m_2}r_{j_{l}}^{m_2}}\right|&\le\frac1{2(h_{m_1}+1)^3} \;\;,\;\;\;\;l=0,\dots,h_{m_1}\\ \left|B_{r_{j_{h_{m_1}+2}}^{m_2}r_{j_{h_{m_1}+1}}^{m_2}}\right|&\le\frac1{2(h_{m_1}+1)^3}\;. \end{align*} Thus we associate step $2$ to the pair $\left(\mathfrak S_{m_1},\widetilde{\Pi}_{m_2}\right)$.
Going on with this construction, we have \begin{align*} &{Step 2N}\leadsto \left(\mathfrak S_{m_{2N-1}},\widetilde{\Pi}_{m_{2N}}\right)\\ &{Step 2N+1}\leadsto \left(\Pi_{m_{2N}},\widetilde{\mathfrak S}_{m_{2N+1}}\right). \end{align*} Now it is well know that given a converging sequence (and it is known that $\{M_{ts}^{\Pi_n}\}_{n\ge1}$, up to passing to a subsequence, is such), every its subsequence converges to the same limit. Since we want to prove that $$ \left|M_{ts}^{\Pi_n}-M_{ts}^{\mathfrak S_n}\right|\stackrel{n\to+\infty}{\longrightarrow}0 $$ we can work with any subsequence of $\{M_{ts}^{\Pi_n}\}_{n\ge1}$ and $\{M_{ts}^{\mathfrak S_n}\}_{n\ge1}$. So, let us prove that $$ \left|M_{ts}^{\Pi_{m_{2N}}}-M_{ts}^{\mathfrak S_{m_{2N+1}}}\right|\stackrel{N\to+\infty}{\longrightarrow}0. $$
Now, observing that $k_m,h_m\ge m$ and $m_N\ge N$, we have \begin{align*} \left|M_{ts}^{\Pi_{m_{2N}}}-M_{ts}^{\mathfrak S_{m_{2N+1}}}\right| &\le\left|M_{ts}^{\Pi_{m_{2N}}}-M_{ts}^{\widetilde{\mathfrak S}_{m_{2N+1}}}\right| +\left|M_{ts}^{\widetilde{\mathfrak S}_{m_{2N+1}}}-M_{ts}^{\widetilde{\Pi}_{m_{2(N+1)}}}\right| +\left|M_{ts}^{\widetilde{\Pi}_{m_{2(N+1)}}}-M_{ts}^{\mathfrak S_{m_{2N+1}}}\right|\\ &\le\frac1{(k_{m_{2N}}+1)^2} +\left|M_{ts}^{\widetilde{\mathfrak S}_{m_{2N+1}}}-M_{ts}^{\widetilde{\Pi}_{m_{2(N+1)}}}\right| +\frac1{(h_{m_{2N+1}}+1)^2}\\ &\le\frac1{(2N+1)^2} +\frac1{(2N+2)^2} +\left|M_{ts}^{\widetilde{\mathfrak S}_{m_{2N+1}}}-M_{ts}^{\widetilde{\Pi}_{m_{2(N+1)}}}\right|. \end{align*}
but here I am stuck, because in order to repeat the argument for the last summand I should pass to a suitable subsequence wrt to $N$ in the whole inequality $$ \left|M_{ts}^{\Pi_{m_{2N}}}-M_{ts}^{\mathfrak S_{m_{2N+1}}}\right| \le\frac1{(2N+1)^2} +\frac1{(2N+2)^2} +\left|M_{ts}^{\widetilde{\mathfrak S}_{m_{2N+1}}}-M_{ts}^{\widetilde{\Pi}_{m_{2(N+1)}}}\right| $$ but I don't know how to do it in a rigorous way.
Moreover, even doing this, and getting something like $$ \left|M_{ts}^{\Pi_{m_{2N}}}-M_{ts}^{\mathfrak S_{m_{2N+1}}}\right| \le\sum_{k\ge N}a_k+\left|M_{ts}^{\Pi_{m_{X}}}-M_{ts}^{\mathfrak S_{m_{Y}}}\right| $$ where $\sum_ka_k$ is a convergent series, how can we control the last summand?