Let $B_1,B_2 \in \mathbb{R}^{n}$ be two linearly independent vectors and $A \in \mathbb{R}^{n\times n}$. Take $B_1$ and multiply it by $A$ until the resulting vector is linearly dependent on the previous. One obtains the following set of linear independent vectors: $$ \{B_1, A\cdot B_1, ..., A^{r_1-1}\cdot B_1\}$$ Let $S_1$ denote the subspace generated by them: $$ S_1 = \langle B_1, ..., A^{r_1-1}\cdot B_1 \rangle$$ we have $A^{r_1} \cdot B_1 \in S_1$. Then assuming $B_2$ does not belong to the subspace generated by the above vectors, consider the following subspace: $$ \bar{S}_1 = \langle B_1, ..., A^{r_1-1}\cdot B_1, B_2, ... , A^{r2-1}\cdot B_2\rangle$$ such that $A^{r_2}\cdot B_2 \in \bar{S}_1$
Question 1
Assume $r_2 \leq r_1$ Is it true, or possible to prove that actually $$ A^{r_2} \cdot B_2 \in \langle B_1, ..., A^{r_2-1}\cdot B_1, A^{r_2} \cdot B_1, B_2, ..., A^{r_2-1}\cdot B_2\rangle$$ that is $A^{r_2}\cdot B_2$ does not depend on higher powers of $A$? We can eventually assume that $r_1 + r_2 = n$ ...
A sketch of my thoughts
Consider the set $$ S_k = \{B_1, ..., A^{k-1}\cdot B_1\} \cup \{B_2, ..., A^{k-1}\cdot B_2\}$$ The vectors in this set are linear dependent for $k > \frac{n}{2}$ while the vectors in the set $$ S_{r_2} = \{B_1, ..., A^{r_2-1}\cdot B_1\} \cup \{B_2, ..., A^{r_2-1}\cdot B_2\}$$ are linear independent. Let $\underline{k} \in [r_2, \lceil\frac{n}{2}\rceil] \cap \mathbb{N}$ be the lowest number such that $S_{\underline{k}}$ contains linear dependent vectors. We should prove that $\underline{k} = r_2+1$ Moreover, it can be safely assumed, that $\underline{k} \in (r_2, r_1) \cap \mathbb{N}$. Therefore $\underline{k} \in [r_2, \min\{r_1,\lceil\frac{n}{2}\rceil\}]\cap \mathbb{N}$
It can be easily be shown that such a $\underline{k}$ has the property that for all $k \geq \underline{k}$ the set $S_k$ contains linearly dependent vectors, and for all $k < \underline{k}$ the set $S_k$ contains linearly independent vectors.
The answer to the Question 1 is no. Here is a counterexample. Take $n=4$ and let's denote by $e_i$ the i$^{\text{th}}$ vector of the canonical basis of $\Bbb R^4$. If $B_1=e_1$, $B_2=e_4$ and if $$AB_1=e_2, \;A^2B_1=e_3,\;A^3B_1=e_1$$ and $AB_2=e_3$ then $$r_1=3,\;r_2=1<r_1$$ but $\langle B_1, AB_1, B_2\rangle$ doesn't contain $AB_2=e_3$.
Edit: The matrix A is $$\begin{bmatrix} 0&0&1&0\\ 1&0&0&0\\ 0&1&0&1\\ 0&0&0&0 \end{bmatrix}$$