There is no need for an epsilon transition, it is a waste of space.
For a regular expression $E$, with resulting automaton $A$, must respect these properties of the transition function, $\delta$:
$A$ has exactly one initial state $q_0$, which is not accessible$^{*}$ from any state. That is,
$$ \delta(q, a) \neq q_0 \quad \forall q \in A, \forall a \in \Sigma$$
$A$ has exactly one final state $q_f$, which is not co-accessible$^{**}$ to any state. That is,
$$ \delta(q_f, a) = \emptyset \quad \forall a\in \Sigma$$
These two points are important. The initial state has no transitions into it, and the final state has no transitions out of it. This means we have no "merge conflicts", as it were.
Let $F$ be the set of all states co-accessible to the final state of $N(s)$, $s_f$.
Let $T$ be the set of all states accessible from the initial state of $N(t)$, $t_0$.
We then have the two states $s_f$ and $t_0$ and their transitions like so:
$$
\begin{align}
F \rightarrow & s_f \rightarrow \emptyset \\
\emptyset \rightarrow &t_0 \rightarrow T \\
\end{align}
$$
You can see how merging the two creates no conflicts or discrepancies:
$$
\begin{align}
F \cup \emptyset \rightarrow & s_ft_0 \rightarrow \emptyset \cup T \\
F \rightarrow & s_ft_0 \rightarrow T \\
\end{align}
$$
$^{*}$ A state $q$ of $A$ is accessible from a state $p$ if there is a computation in $A$ whose source is $p$ and whose destination is $q$. A state $q$ is accessible if it is accessible from an initial state.
$^{**}$ A state $p$ of $A$ is co-accessible to a state $q$ if there is a computation in $A$ whose source is $p$ and whose destination is $q$. A state $p$ is co-accessible if it is co-accessible to a final state.