Given $ A=\begin{pmatrix} A_{11} & A_{12} \\ A_{21} & A_{22}\\ \end{pmatrix}$, a $2n\times 2n$ real matrix with each $A_{ij}$, a $n\times n$ matrix, take $A_1=\begin{pmatrix} A_{11} & 0 \\ 0 & A_{22}\\ \end{pmatrix}$, $ A_2=\begin{pmatrix} 0 & A_{12} \\ A_{21} & 0\\ \end{pmatrix}$, then when is $$Rank(A)\leq Rank(A_1)+Rank(A_2)\leq 2\max(Rank(A_1),Rank(A_2))\leq 4Rank(A)?$$
What about if we deal with just $0-1$ matrices?
Typically could we expect relation between $Rank(A)$,$Rank(A_i)$,$Rank(A_1)+Rank(A_2)$?
Does $$Rank(A)\leq Rank(A_1)+Rank(A_2)\leq 2\max(Rank(A_1),Rank(A_2))\leq 4Rank(A)?$$ hold over matrices from $\Bbb F_p$ (including $0/1$ case from $\Bbb F_p$)?
The inequality $$ rank(A) \leq \max (rank(A_1), rank(A_2)) $$ is false, even for $0$-$1$ matricies. A counterexample is given by $$ A = \left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1\end{array} \right), $$ which has $rank(A) = 4$ but $rank(A_1) = rank(A_2) = 2$.
However, it is true that $$ rank(A) \leq rank(A_1) + rank(A_2), $$ which follows from the fact that $$ image(A) \subseteq image(A_1) + image(A_2), $$ which should be clear. (In fact this shows that this inequality holds whenever $A = A_1 + A_2$ regardless of the special form of $A_1$ and $A_2$ you've supposed.)
I think the example above suggests that even among $0$-$1$ matricies and with your special assumptions on $A_1$ and $A_2$ this is the best you can hope for.
Your right hand inequality, $$ \max (rank(A_1), rank(A_2)) \leq 2 \; rank(A), $$ is also true (in your specific setup). To see this, let $u_1, \dots, u_k$ be a basis for the image of $A$. For each $i$, let $v_i$ be the projection of $u_i$ onto the first $n$ components, and let $w_i$ be the projection onto the last $n$. (In particular $u_i = v_i + w_i$.)
Then it's easy to see that $\lbrace v_i, w_i \rbrace_{1 \leq i \leq k}$ spans the image of both $A_1$ and $A_2$. (For instance, suppose $a_1$ is the first column of $A$ and we can write $a_1 = \sum c_i u_i$. Then the first column of $A_1$ can be written $\sum c_i v_i$, and the first column of $A_2$ can be written $\sum c_i w_i$.)