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I doubt that the condition $\mathrm{End}_{A_0}(M_0)=k$ is enough. I address the problem in some special cases here by adding the following hypothesis:

1. The $A/(t-x)A$-module $M/(t-x)M$ is irreducible for generic $x\in k$ and that $A\to \mathrm{End}_{k[t]}(M)$ is injective (thus we regard $A$ as a subalgebra).
2. Every subquotient $N$ of $M_0$ satisfies $\mathrm{End}_{A_0}(N)=k$.

First, we study the central fibers. We denote $\bar M = M_0$ and $\bar A=$ the image of $A$ in $\mathrm{End}_k(\bar M)$

• For $1\le i < j \le r$, we have $\bar A^{ji}\bar M \subseteq e_jF_i = \bigoplus_{l \le i} e_je_l\bar M = 0$. So $\bar A^{ji} = 0$ as the module $\bar M$ is faithful.
• For $1\le i=j\le r$, the $\bar A^{ii}$-module $\bar M^i$ is irreducible. By Jacobson density theorem, $\bar A^{ii}\cong \mathrm{End}_k(\bar M^i)$.
• For $1\le j < i\le r$, the space $\mathrm{Hom}_k(\bar M^i, \bar M^j)$ has a natural $\bar A^{jj}$-$\bar A^{ii}$-bimodule structure, which is irreducible since $\bar M^i$ and $\bar M^j$ are. We claim that $\bar A^{ji} = \mathrm{Hom_k}(\bar M^i, \bar M^j)$. Suppose otherwise. We would have $\bar A^{ji} = 0$ by the irreducibility. Let $i$ be minimum with this property. Therefore $\bar A^{j, i-1} \cong \mathrm{Hom}_k(\bar M^{i-1}, \bar M^j)$. The multiplication $\bar A^{j, i-1}\otimes_k \bar A^{i-1, i}\to \bar A^{j, i}$ considered as restriction of $\mathrm{Hom}_k(\bar M^{i-1}, \bar M^{j})\otimes_k \mathrm{Hom}_k(\bar M^{i}, \bar M^{i-1})\to \mathrm{Hom}_k(\bar M^{i}, \bar M^{j})$ forces also $A^{i-1, i}_0=0\subseteq \mathrm{Hom}_k(\bar M^{i}, \bar M^{i-1})$. However, this implies that $F_i/F_{i-2}\cong e_i(F_i/F_{i-2})\oplus e_{i-1}(F_i/F_{i-2})$ as $\bar A$-module, contradicting the hypothesis that $\mathrm{End}_{\bar A}(F_i/F_{i-2})=k$.

• By 5. and Nakayama lemma we see that $\hat A^{ji}=\mathrm{Hom}_{k[[t]]}(\hat M^i, \hat M^j)$ for $j \le i$.
• By hypothesis, $\hat M[t^{-1}]$ is irreducible, so Jacobson density theorem gives $\hat A[t^{-1}] = \mathrm{End}_{k((t))}(\hat M[t^{-1}])$. Thus $\hat A$ is a full $k[[t]]$-lattice in $\mathrm{End}_{k[[t]]}(\hat M)[t^{-1}]$. There exists therefore $N\in \mathbb{N}$ such that $t^N \mathrm{End}_{k[[t]]}(\hat M)\subseteq \hat A$. Thus, for $j > i$, repeating the arguments of bimodules in 5., we have $\hat A^{ji} = t^{a_{ji}}\mathrm{Hom}_{k[[t]]}(\hat M^i, \hat M^j)$ for some $0 < a_{ji} \le N$.

8. Similarly, write $\hat B^{ji} = d_j\hat Bd_i$. For any $1\le i , j\le r$, we proceed as in 7., so that $\hat B^{ji} = t^{b_{ji}}\mathrm{Hom}_{k[[t]]}(\hat M^i, \hat M^j)$ for some $b_{ji} \in \mathbb{Z}$ and $b_{ji} \le a_{ji}$. The finiteness of $\hat B$ over $k[[t]]$ gives some bounds on these integers:

I believe that the first hypothesis can be easily elimnated, but the second one should be essential.

I doubt that the condition $\mathrm{End}_{A_0}(M_0)=k$ is enough. I address the problem in some special cases here by strengthening the hypothesis on endomorphisms:

• Every subquotient $N$ of $M_0$ satisfies $\mathrm{End}_{A_0}(N)=k$.

Proof.

By replacing $A$ (resp. $B$) with its image in $\mathrm{End}_{k[t]}(M)$ (resp. $\mathrm{End}_{k[t]}(M)[t^{-1}]$), we regard it as a $k[t]$-subalgebra.

First, we study the central fibers. We denote $\bar M = M_0$ and $\bar A=$ the image of $A$ in $\mathrm{End}_k(\bar M)$

• For $1\le i < j \le r$, we have $\bar A^{ji}\bar M \subseteq e_jF_i = \bigoplus_{l \le i} e_je_l\bar M = 0$. So $\bar A^{ji} = 0$ as the module $\bar M$ is faithful.
• For $1\le i=j\le r$, the $\bar A^{ii}$-module $\bar M^i$ is irreducible. By Jacobson density theorem, $\bar A^{ii}\cong \mathrm{End}_k(\bar M^i)$.
• For $1\le j < i\le r$, the space $\mathrm{Hom}_k(\bar M^i, \bar M^j)$ has a natural $\bar A^{jj}$-$\bar A^{ii}$-bimodule structure, which is irreducible since $\bar M^i$ and $\bar M^j$ are. We claim that $\bar A^{ji} = \mathrm{Hom_k}(\bar M^i, \bar M^j)$. Suppose otherwise. We would have $\bar A^{ji} = 0$ by the irreducibility. Let $i$ be minimum with this property. Therefore $\bar A^{j, i-1} \cong \mathrm{Hom}_k(\bar M^{i-1}, \bar M^j)$. The multiplication $\bar A^{j, i-1}\otimes_k \bar A^{i-1, i}\to \bar A^{j, i}$ considered as restriction of $\mathrm{Hom}_k(\bar M^{i-1}, \bar M^{j})\otimes_k \mathrm{Hom}_k(\bar M^{i}, \bar M^{i-1})\to \mathrm{Hom}_k(\bar M^{i}, \bar M^{j})$ forces also $\bar A^{i-1, i}=0\subseteq \mathrm{Hom}_k(\bar M^{i}, \bar M^{i-1})$. However, this implies that $F_i/F_{i-2}\cong e_i(F_i/F_{i-2})\oplus e_{i-1}(F_i/F_{i-2})$ as $\bar A$-module, contradicting the hypothesis that $\mathrm{End}_{\bar A}(F_i/F_{i-2})=k$.

• For any $0\le i, j\le r$, since for each $s\in \mathbb{Z}$, the succesive quotient $t^s\mathrm{Hom}_{k[[t]]}(\hat M^i, \hat M^j)/t^{s+1}\mathrm{Hom}_{k[[t]]}(\hat M^i, \hat M^j)$ is isomorphic as $\bar A^{jj}-\bar A^{ii}$-bimodule to $\mathrm{Hom}_{k}(\bar M^i, \bar M^j)$, we have $\hat A^{ji} = t^{a_{ji}}\mathrm{Hom}_{k[[t]]}(\hat M^i, \hat M^j)$ for some $a_{ji}\in \mathbb{N}\cup \{+\infty\}$ (where $t^{+\infty}$ means $0$).
• By 5. and Nakayama lemma we see that $a_{ji}=0$ for $j \le i$.

8. Similarly, write $\hat B^{ji} = d_j\hat Bd_i$. For any $1\le i , j\le r$, we proceed as in 7., so that $\hat B^{ji} = t^{b_{ji}}\mathrm{Hom}_{k[[t]]}(\hat M^i, \hat M^j)$ for some $b_{ji} \in \mathbb{Z}\cup \{+\infty\}$ and $b_{ji} \le a_{ji}$. The finiteness of $\hat B$ over $k[[t]]$ gives some bounds on these exponents:

I believe that the strengthened hypothesis is essential. We notice also that the extension of $M_0$ into a $B_0$-module may not be unique. I guess the uniqueness corresponds to the case where $b_{ji}=0$ (minimum possible) for those $b_{ji}\neq a_{ji}$.

6. Lift the idempotents $e^1, \cdots, e^r$ to idempotents $d^1, \cdots, d^r\in \hat A$, which is possible by the $t$-adic completeness. By modifying $d'_1=d_2,\; d'_2=(1-d'_1)d_2(1-d'_1),\; d'_3=(1-d'_1)(1-d'_2)d_3(1-d'_1)(1-d'_2)$, etc, we can make $d_id_j = 0$ for $i\neq j$.

7. Define as before $\hat A^{ji}= d_j\hat Ad_i$ and $\hat M^i = d_i\hat M$. Then

6. Lift the idempotents $e^1, \cdots, e^r$ to idempotents $d^1, \cdots, d^r\in \hat A$, which is possible by the $t$-adic completeness. By modifying $d'_1=d_1,\; d'_2=(1-d'_1)d_2(1-d'_1),\; d'_3=(1-d'_1)(1-d'_2)d_3(1-d'_1)(1-d'_2)$, etc, we can make $d_id_j = 0$ for $i\neq j$.

7. Define as before $\hat A^{ji}= d_j\hat Ad_i$ and $\hat M^i = d_i\hat M$. Then

• By 5. and Nakayama lemma we see that $\hat A^{ji}=\mathrm{Hom}_{k[[t]]}(\hat M^i, \hat M^j)$ for $j \le i$.
• By hypothesis, $\hat M[t^{-1}]$ is irreducible, so Jacobson density theorem gives $\hat A[t^{-1}] = \mathrm{End}_{k((t))}(\hat M[t^{-1}])$. Thus $\hat A$ is a full $k[[t]]$-lattice in $\mathrm{End}_{k[[t]]}(\hat M)$. There exists therefore $N\in \mathbb{N}$ such that $t^N \mathrm{End}_{k[[t]]}(\hat M)\subseteq \hat A$. Thus, for $j > i$, repeating the arguments of bimodules in 5., we have $\hat A^{ji} = t^{a_{ji}}\mathrm{Hom}_{k[[t]]}(\hat M^i, \hat M^j)$ for some $0 < a_{ji} \le N$.

• By 5. and Nakayama lemma we see that $\hat A^{ji}=\mathrm{Hom}_{k[[t]]}(\hat M^i, \hat M^j)$ for $j \le i$.
• By hypothesis, $\hat M[t^{-1}]$ is irreducible, so Jacobson density theorem gives $\hat A[t^{-1}] = \mathrm{End}_{k((t))}(\hat M[t^{-1}])$. Thus $\hat A$ is a full $k[[t]]$-lattice in $\mathrm{End}_{k[[t]]}(\hat M)[t^{-1}]$. There exists therefore $N\in \mathbb{N}$ such that $t^N \mathrm{End}_{k[[t]]}(\hat M)\subseteq \hat A$. Thus, for $j > i$, repeating the arguments of bimodules in 5., we have $\hat A^{ji} = t^{a_{ji}}\mathrm{Hom}_{k[[t]]}(\hat M^i, \hat M^j)$ for some $0 < a_{ji} \le N$.