I don’t see how to justify the statement in question, but I do see how to justify a weaker statement that is still sufficient to prove the theorem.
Suppose that $G-\{x_1,x_2,\ldots,x_k\}$ is connected and and $\deg v_i>k$ for each $i\in\{1,\ldots,k\}$. Then each $v_i$ has at least $k+1$ neighbors in $G$, and at most $k-1$ of them can be in $\{x_1,x_2,\ldots,x_k\}$, so there must be at least two vertices $u_i$ and $v_i$ in $G-\{x_1,x_2,\ldots,x_k\}$ that are neighbors of $x_i$ in $G$. At least one of $u_i$ and $v_i$ must be different from $y_i$; say $u_i\ne y_i$. I claim that this implies that $G-\{x_1y_1,\ldots,x_ky_k\}$ is connected, which is a contradiction.
Let $v$ and $w$ be any two vertices of $G$. If $v,w\notin\{x_1,\ldots,x_k\}$, then there is a path from $v$ to $w$ in $G-\{x_1,\ldots,x_k\}$ and hence in $G-\{x_1y_1,\ldots,x_ky_k\}$. If $v=x_i$ and $w\notin\{x_1,\ldots,x_k\}$, there is a path $P$ from $w$ to $u_i$ in $G-\{x_1y_1,\ldots,x_ky_k\}$, and we can add the edge $u_ix_i$ to $P$ to get a path from $w$ to $x_i$ in $G-\{x_1y_1,\ldots,x_ky_k\}$. Finally, if $v=x_i$ and $w=x_j$ with $i\ne j$, there is a path $P$ from $u_i$ to $u_j$ in $G-\{x_1y_1,\ldots,x_ky_k\}$, and we can extend it with the edges $x_iu_i$ and $u_jx_j$ to get a path from $x_i$ to $x_j$ in $G-\{x_1y_1,\ldots,x_ky_k\}$. This proves the claim.
This contradiction shows that there must be at least one $i\in\{1,\ldots,k\}$ such that $\deg x_i\le k$, and since we already know that $k=\lambda(G)\le\delta(G)\le\deg x_i$, we must have $\deg x_i=k$. Deleting the neighbors of $x_i$ disconnects $G$, so $\kappa(G)\le k=\lambda(G)$.
Without loss of generality, assume that $m \le n$. And we'd like to show that the vertex-connectivity $\kappa(G) = m-1$, where $G$ is the complete bipartite graph on $m + n$ vertices with one edged, say $(a,b)$, removed.
It is enough to show that removing $m-2$ vertices cannot make $G$ disconnected. Suppose the left and the right partite set are of size $m, n$ respectively. After removing $m-2$ vertices from $G$, we get a complete bipartite graph with at most one edge removed. Suppose the partite sets are $A$ and $B$ of the resulting graph. We also know that $|A|, |B| \ge 2$.
Take any two vertices $u$ from $A$ and $v$ from $B$. As $|A|, |B| \ge 2$, we can find $u'\in A\setminus \{a\}$, and $v'\in B\setminus \{b\}$. Thus $u\sim v'\sim u' \sim v$, and so $A$ and $B$ are connected.
Best Answer
The result that $\delta(G) \ge \frac{n+1}{4} \implies \kappa'(G) = \delta(G)$ for bipartite graphs follows from Theorem 3.1 in the paper you cited.
Let $P \cup Q$ be the bipartition of $G$ with $|P| \ge \frac n2$ and $|Q| \le \frac n2$. Then $\delta(G) > \frac12|Q|$, so any two vertices $x,y \in P$ have a common neighbor in $Q$: condition (ii) of Theorem 3.1 holds.
On the other hand, $\delta(G) \ge \frac n4$ is not enough to conclude that $\kappa'(G) = \delta(G)$, even when $G$ is bipartite. As mentioned in the same paper, $G$ could be the union of two disjoint copies of $K_{n/4,n/4}$, with $\delta(G)=\frac n4$ but $\kappa'(G)=0$.