Given a bipartite graph $G$ with parts $A_i$ and $B_i$ with maximum degree $d$. We may assume that $|U|,|V|\le n$ and $n\ge d$.
I am wondering is it possible to decompose $G$ into vertex-disjoint bipartite graphs $G_i$ with parts $A_i, B_i$ such that $|A_i|, |B_i|=d^{O(1)}$, assuming that $d$ is fixed and $n$ tends to infinity, and $$\left|E(G)\setminus\bigcup_i E(G_i)\right| = d^{O(1)}?$$
In general, if $G$ is allowed to be an arbitrary bipartite graph with maximum degree $d$, and $d$ is allowed to be an arbitrary fixed integer, the answer is NO. In fact, if $G$ is any connected bipartite graph on $n$ vertices and with maximum degree $d$ and each $G_i$ has $d^{O(1)}$ vertices, the condition $$\left|E(G)\setminus\bigcup_i E(G_i)\right| = d^{O(1)}$$ is never met, provided $n$ is large enough relative to $d$ more precisely $n > (d^{O(1)})^2+1$.
Indeed, let $G$ be a connected bipartite graph on $n$ vertices and with maximum degree $d$, and assume $n$ is large enough relative to $d$. Then contract each $G_i$ to a single vertex $v_i$ and call the resulting graph $G'$. Then, as $G$ is connected then so must $G'$ be connected. So we now show that the bound on the number of edges between the $G_i$s forces $G'$ to be not connected however. To this end, $G'$ has $n/d^{O(1)} >> d^{O(1)}$ vertices, as each $G_i$ has only $d^{O(1)}$ vertices. However, the number of edges in $G'$ is only $\left|E(G)\setminus\bigcup_i E(G_i)\right|$ $=$ $d^{O(1)}$. Thus, $G'$ has many more vertices [$n/d^{O(1)}$ vertices] than edges [only $d^{O(1)}$ edges] so it is impossible for $G'$ to be connected after all. Thus, as $G'$ is not connected, the graph $G$ cannot be connected either, which contradicts the given that $G$ is connected.