Explicit description of exponentials of étalé spaces It is well known that the category $\mathit{Sh}(X)$ of sheaves of sets on a topological space $ X $ is a topos.
On the other hand, there exists a natural equivalence of categories between $\mathit{Sh}(X)$ and the category $\mathit{Et}(X)$ of étalé spaces over  $ X $, so that the category of etale spaces over $ X $ is a topos, too.
The constructions in $\mathit{Sh}(X)$ are translatable to $\mathit{Et}(X)$, and vice versa.
For example, products of sheaves correspond to fiber products of étalé spaces.
The description of the exponentials in $\mathit{Sh}(X)$ is given, e.g., in [Mac Lane and Moerdijk, Sheaves in Geometry and Logic].  The question that arises is


What is the explicit construction of exponentials in $Et(X)$?
 A: Let $[E,E’]$ be the internal hom of two étale spaces $\pi : E \to X$ and $\pi’: E’ \to X$. I will write $\mathrm{Hom}_X(A,B)$ for the set of morphisms of étale spaces $A$ and $B$ over $X$, and $A \times_X B$ for the product of $A$ and $B$ as étale spaces over $X$.
To compute the sections of $[E,E’]$, you can use
$$\mathrm{Hom}_X(U,[E,E’]) \simeq  \mathrm{Hom}_X(U \times_X E, E’) \simeq \mathrm{Hom}_X(U \times_X E, U \times E’).$$
The first isomorphism uses the universal property of the exponential, the second isomorphism uses that the image of $U \times E$ is always contained in $(\pi’)^{-1}(U) =U \times E’$, so there is a unique factorization through $U \times E’$.
The elements of the total space correspond to stalks, so the fiber of $[E,E’]$ above $x \in X$ is given by the set
$$[E,E’]_x \simeq \varinjlim_i \mathrm{Hom}_X(U_i \times_X E, U_i \times_X E’)$$
where $(U_i)_{i \in I}$ is a projective system of open neighborhoods of $x$ such that every open neighborhood of $x$ contains some $U_i$ (for example, if $X$ is a metric space, you can take open balls of shrinking radius).
A basis of open sets on $[E,E’] = \bigsqcup_x [E,E’]_x$ is given by the sets $A_{(U,s)} = \{ s_x : x \in U \}$, for $U$ in some basis of open sets of $X$, $s \in \mathrm{Hom}_X(U \times_X E, U \times_X E’)$, and $s_x$ the element of $[E,E’]_x$ corresponding to $s$. This topology is often not easy to visualize. The spaces $E$ and $E’$ are already typically non-Hausdorff, and even if they are Hausdorff there is no guarantee that $[E,E’]$ will be as well. But maybe it helps to keep in mind that the subspace topology on the fibers is always the discrete topology (as for any étale space).
For example, take $X= \mathbb{R}$, $E = \mathbb{R}-\{0\}$ and $E’ = \mathbb{R}\sqcup\mathbb{R}$ with $\pi$ and $\pi’$ the natural projections.
We first look at $x \neq 0$. For a small open interval $U$ containing $x\neq 0$, there are exactly two maps $E \times_X U \to E’ \times_X U$ (because $E \times_X U \simeq U$ and $E’ \times_X U \simeq  U \sqcup U$).
However, if $U$ is a small open interval containing $x=0$, then $E \times_X U = U -\{0\}$ has two components, and because of this there are four maps to $E’ \times_X U \simeq (U-\{0\}) \sqcup (U - \{0\})$.
So in this example, $[E,E’]$ has four points above $x=0$ and two points above each $x \neq 0$. Further, $[E,E’]$ has four global sections corresponding to the four continuous maps $E \to E’$ over $X$. This gives four open sets $A_{(X,s)}$ in $[E,E’]$, with each of these homeomorphic to $\mathbb{R}$. But these four lines are glued together to a single line on $\{ x \in \mathbb{R} : x > 0\}$ iff the corresponding sections agree on $\{ x \in \mathbb{R} : x > 0\}$, and similarly over $\{x \in \mathbb{R} : x < 0 \}$.
Here is a picture:

The gluing of different lines is supposed to be “instant” (just as in the case of the line with two origins) and as a result the space is non-Hausdorff.
A: Depending on what you mean by "explicit", the following answer will be satisfying or not:

Let $F : {\sf Et}(X) \leftrightarrows {\sf Sh}(X) : G$ be the equivalence in subject. Let $[-,-]$ be the exponential of sheaves, i.e. on presheaves, i.e. the sheaf $U\mapsto {\sf Sh}(X)(yU \times A,B)$ (if I remember well, it is enough that $B$ is a sheaf for this to be a sheaf). Then, the exponential $\langle -,-\rangle$ on étale spaces is the functor $(E,E')\mapsto G[FE, FE']$.

Indeed, let's prove that ${\sf Et}(X)(E\times E', E'')\cong {\sf Et}(X)(E, \langle E',E''\rangle)$:
$$
\begin{align*}
{\sf Et}(X)(E\times E', E'') & \cong {\sf Sh}(X)(F(E\times E'), FE'') \\
& \cong {\sf Sh}(X)(FE\times FE', FE'')\\
& \cong {\sf Sh}(X)(FE, [FE', FE''])\\
& \cong {\sf Et}(X)(E, G[FE', FE'']) \\
& = {\sf Et}(X)(E, \langle E', E''\rangle).
\end{align*}
$$
I have used the following facts: 1. $F$ preserves products (and for that matter, all limits: it is an equivalence). 2. ${\sf Sh}(X)$ is cartesian closed.
