Complexity of finding Exact Size Cut-Sets in Bipartite Graphs
This is a partial answer that proves the NP-completeness of your problem, by a reduction from the EXACT-CUT problem for general graphs1. 

Given an instance $(G=(V,E),k)$ of the EXACT-CUT problem (that asks whether there is a cut of size $k$ in the graph $G$), we construct a new graph $G'$ as follows.


For each vertex $v\in V$, construct vertices $v_0^0,\ldots,v_m^0,v_0^1,\ldots,v_m^1$ as well as edges $(v_i^0, v_j^1)$ for all $i,j\in\{0,\ldots,m\}$ (so that they form a biclique). Denote $V_v=\{v_0^0,\ldots,v_m^0,v_0^1,\ldots,v_m^1\}$.
For each edge $(u,v)\in E$, construct two edges $(u_0^0, v_0^1)$ and $(u_0^1, v_0^0)$.


Here $m=2k$. Note $G'$ is a bipartite graph.

Now we can see if $(A,B)$ is a cut of size $k$ in $G$, then $(\cup_{v\in A}V_v,\cup_{v\in B}V_v)$ is a cut of size $2k$ in $G'$.

On the other hand, suppose there is a cut $(A',B')$ of size $2k$ in $G'$. Now consider a vertex $v$. Assume $|A'\cap\{v_0^0,\ldots,v_m^0\}|=a$ and $|A'\cap\{v_0^1,\ldots,v_m^1\}|=b$, then we have
\begin{align}
m=2k&amp;\ge a(m+1-b)+b(m+1-a)\\
&amp;=(m+1)(a+b)+\frac{1}{2}(a-b)^2-\frac{1}{2}(a+b)^2\\
&amp;\ge(m+1)(a+b)-\frac{1}{2}(a+b)^2,
\end{align}
which implies $a+b&lt;1$ or $a+b&gt;2m+1$, i.e. $a=b=0$ or $a=b=m+1$. This means all vertices in $V_v$ must belong to the same cut-set. Hence $(\{v\mid V_v\subseteq A'\}, \{v\mid V_v\subseteq B'\})$ is a cut of size $k$ in $G$.

Now we have shown that there is a cut of size $k$ in $G$ if and only if there is a cut of size $2k$ in $G'$, which concludes the proof of the NP-completeness of the EXACT-CUT problem on bipartite graphs.



1 Your reasoning for the NP-completeness of the EXACT-CUT problem is incorrect because you used a Turing reduction while a many-one reduction is required. However, it is not hard to find such a polynomial-time many-one reduction.