You don’t need any empty transitions. I’m working with non-deterministic automata.
Let $N_1=\langle\Sigma,S_1,s_0,\delta_1,A_1\rangle$, where $\Sigma$ is the alphabet, $S_1$ is the state set, $s_0$ is the initial state, $\delta_1:S_1\times\Sigma\to\wp(S_1)$ is the transition function, and $A_1\subseteq S_1$ is the set of acceptor states. Similarly, let $N_2=\langle\Sigma,S_2,t_0,\delta_2,A_2\rangle$. Assume without loss of generality that $S_1\cap S_2=\varnothing$.
Now let $N=\langle\Sigma,S,s_0,\delta,A_2\rangle$, where $S=S_1\cup(S_2\setminus\{t_0\})$ and $\delta:S\times\Sigma\to S$ is defined as follows:
$\delta(s,\sigma)=\begin{cases} \delta_1(s,\sigma),&\text{if }s\in S_1\setminus A_1\\ \delta_2(s,\sigma),&\text{if }s\in S_2\\ \delta_1(s,\sigma)\cup\delta_2(s,\sigma),&\text{if }s\in A_1\;. \end{cases}$
Suppose that $x=\sigma_1\dots\sigma_m\in L(N_1)$. Then there are states $s_1\dots,s_m\in S_1$ such that $s_k\in\delta_1(s_{k-1},\sigma_k)\quad\text{for}\quad k=0,\dots,m\tag{1}$ and $s_n\in A_1$. Similarly, if $y=\tau_1\dots\tau_n\in L(N_2)$, there are states $t_1,\dots,t_n\in S_2$ such that $t_k\in\delta_2(t_{k-1},\tau_k)\quad\text{for}\quad k=0,\dots,n\tag{2}$ and $t_n\in A_2$.
For $k=0,\dots,m$ let $q_k=s_k$, and for $k=m+1,\dots,m+n$ let $q_k=t_{k-m}$. $(1)$ implies that $q_k\in\delta(q_{k-1},\sigma_k)\quad\text{for}\quad k=0,\dots,m\;,\tag{3}$ and $(2)$ implies that $q_k\in\delta(q_{k-1},\tau_{k-m})\quad\text{for}\quad k=m+2,\dots,m+n\;,\tag{4}$ with $q_{m+n}=t_n\in A_2$. Finally, $s_m\in A_1$, so the third part of the definition of $\delta$ shows that $q_{m+1}\in\delta_2(t_0,\tau_1)\subseteq\delta(q_m,\tau_1)\;.\tag{5}$ The word $xy$ is therefore accepted by $N$ with the computation combining $(3),(5)$, and $(4)$.
To finish the job you must show that if $w\in L(N)$, then $w\in L(N_1)\cdot L(N_2)$. Suppose that $w=\sigma_1\dots\sigma_n\in L(N)$. Then there are states $q_0=s_0,q_1,\dots,q_n\in S$ such that $q_n\in A_2$ and $q_k\in\delta(q_{k-1},\sigma_k)\quad\text{for}\quad k=0\dots,n\;.$ Note that $q_n\in A_2\subseteq S\setminus S_1$. Let $m=\min\{k\in\{1,\dots,m\}:q_m\in S\setminus S_1\}$. To finish the argument, show that $q_{m-1}\in A_1$ and $q_m\in\delta_2(t_0,\sigma_m)$, and conclude that $\sigma_1\dots\sigma_{m-1}\in L(N_1)$ and $\sigma_m\dots\sigma_n\in L(N_2)$.