For all $k\ge 0$ we set $a_k(M)=a(H_{R_+}^{k}(M))$.
Lemma. Let $z\in R_1$ be an $M$-filter regular element. Then, for all $k\ge 0$, we have $a_{k+1}(M)+1\le a_k(M/zM)\le\max\{a_k(M),a_{k+1}(M)+1\}.$
Proof. By the definition of $M$-filter regularity and a similar result for modules to this one Noetherian ring and primary decomposition result we get $(0:_Mz)\subseteq\bigcup_{n\ge 0}(0:_MR_+^n)$. In particular $(0:_Mz)$ is annihilated by $R_+^n$ for some $n\ge 0$ which in terms of local cohomology means that $H_{R_+}^i(0:_Mz)=0$ for all $i\ge 1$. This shows easily that $H_{R_+}^i(M)=H_{R_+}^i(M/(0:_Mz))$ for all $i\ge 1$. From the following short exact sequence $0\longrightarrow M/(0:_Mz)\stackrel{z\cdot}\longrightarrow M\longrightarrow M/zM\longrightarrow 0$ we get a long exact sequence for the local cohomology $H_{R_+}^i(M)_n\longrightarrow H_{R_+}^i(M/zM)_n\longrightarrow H_{R_+}^{i+1}(M)_{n-1}\longrightarrow H_{R_+}^{i+1}(M)_n$ and this proves the assertion.
Proposition. Let $z_1,...,z_s\in R_1$ be an $M$-filter regular sequence. Then $\max\{a_i(M)+i|\ i=0,\dots,s\}=\text{max}\lbrace a((z_1,...,z_i)M:R_{+}/(z_1,...,z_{i})M)|\ i=0,...,s\rbrace.$
Proof. By using Lemma successively we get $a_i(M)+i\le a_0(M/(z_1,\dots,z_i)M)\le\max\{a_j(M)+j|\ j=0,\dots,i\}.$ This implies that $\max\{a_i(M)+i|\ i=0,\dots,t\}=\max\{a_0(M/(z_1,\dots,z_i)M)|\ i=0,\dots,t\}$ for all $t\le s$.
Now we identify $H_{R_+}^{0}(M/(z_1,\dots,z_i)M))$ with $\bigcup_{n\ge 0}(z_1,...,z_i)M:R_{+}^n/(z_1,...,z_{i})M.$ Set $a=a_0(M/(z_1,\dots,z_i)M)$. Then we have $H_{R_+}^{0}(M/(z_1,\dots,z_i)M))_a\subseteq(z_1,...,z_i)M:R_{+}/(z_1,...,z_{i})M\subseteq H_{R_+}^{0}(M/(z_1,\dots,z_i)M))$ and therefore $a((z_1,...,z_i)M:R_{+}/(z_1,...,z_{i})M)=a$.
Edit. $H_{R_+}^{0}(M/(z_1,\dots,z_i)M))_a\subseteq(z_1,...,z_i)M:R_{+}/(z_1,...,z_{i})M$ since $H_{R_+}^{0}(M/(z_1,\dots,z_i)M))$ is a graded $R$-module and $H_{R_+}^{0}(M/(z_1,\dots,z_i)M))_a$ is its top graded part, so $R_+H_{R_+}^{0}(M/(z_1,\dots,z_i)M))_a=0$.
$(z_1,...,z_i)M:R_{+}/(z_1,...,z_{i})M\subseteq H_{R_+}^{0}(M/(z_1,\dots,z_i)M))$ implies $((z_1,...,z_i)M:R_{+}/(z_1,...,z_{i})M)_j\subseteq H_{R_+}^{0}(M/(z_1,\dots,z_i)M))_j$ for all $j$. For $j>a$ the right hand side is $0$, while for $j=a$ is not. This shows that $a((z_1,...,z_i)M:R_{+}/(z_1,...,z_{i})M)=a$.