Definition 1
Let $\mathcal F$ be a presheaf of sets(or abelian groups or rings, etc.) on a toplogical space $X$.
Let $Et(\mathcal F)$ be the disjoint union $\cup_{x \in X} \mathcal F_x$.
Let $U$ be an open subset of X.
Let $s \in \mathcal F(U)$.
We denote by $[U, s]$ the subset {$s_x; x \in U$} of $Et(\mathcal F)$.
Let Open$(X)$ be the set of open subsets of $X$.
We define a topology on $Et(\mathcal F)$ as the one generated by the subset {$[U, s]; U \in$ Open($X), s \in \mathcal F(U)$} of the power set of $Et(\mathcal F)$.
Let $\pi:Et(\mathcal F)\rightarrow X$ be the canonical map which sends a germ $s_x$ to $x$.
Let $\mathcal F^+(U)$ be the set {$f:U \rightarrow Et(\mathcal F)$; $f$ is a continuous map and $\pi f = id_U$}.
Clearly $\mathcal F^+(U)$ defines a sheaf $\mathcal F^+$ on $X$.
Lemma 2
Let $\mathcal F^+(U)$ be as above.
Then $\mathcal F^+(U)$ = {$f:U \rightarrow Et(\mathcal F)$; $f$ is a map such that for each $x \in U$ there exists an open neighborhood $U_x$ of $x$ contained in $U$ and $s \in \mathcal F(U_x)$ such that $f(y) = s_y$ for each $y \in U_x$}.
Proof:Clear.
Definition 3
Let $\mathcal F$ be a presheaf on a toplogical space $X$.
Let $U$ be an open subest of $X$.
Let $s \in \mathcal F(U)$
We define a map $\tilde{s}: U \rightarrow Et(F)$ by $\tilde{s}(x)$ = $s_x$ for each $x \in U$.
Clearly $\tilde{s} \in \mathcal{F^+(U)}$.
Definition 4
Let $\mathcal F$ be a presheaf on a toplogical space $X$.
Let $U$ be an open subest of $X$.
We define a map $\iota_U: \mathcal F(U) \rightarrow \mathcal F^+(U)$ by $\iota_U(s) = \tilde{s}$, where $\tilde{s}$ is defined in Definition 3.
Clearly $\iota_U$'s define a morphism $\iota:\mathcal F \rightarrow \mathcal F^+$.
We call $\iota$ the canonical morphism.
The folowing lemma is fundametal.
Lemma 5
Let $\mathcal F$ be a sheaf on a toplogical space $X$.
Then the canonical morphism $\iota:\mathcal F \rightarrow \mathcal F^+$ is an isomorphism.
Proof:
Let $U$ be an open subest of $X$.
It suffices to prove that $\iota_U: \mathcal F(U) \rightarrow \mathcal F^+(U)$ is an isomorphism.
Let $s$ and $t$ be $\in \mathcal F(U)$.
Suppose $\iota_U(s)$ = $\iota_U(t)$.
This means that $s_x$ = $t_x$ for each $x \in U$.
Hence there exists an open neghborhood $U_x$ of $x$ for exch $x \in U$ such that $s|U_x$ = $t|U_x$.
Since $\mathcal F$ is a sheaf, $s$ = $t$.
Hence $\iota$ is injective.
It remains to prove that $\iota$ is surjective.
Let $\sigma \in \mathcal F^+(U)$.
There exists an open cover $U_i$ of $U$ and $s_i \in \mathcal F(U_i)$ such that $\sigma(x) = s_i(x)$ for each $x \in U_i$.
Since $s_i|U_i \cap U_j = s_j|U_i \cap U_j$ by the above claim, there exists $s \in \mathcal F(U)$ such that $s|U_i$ = $s_i$ for each $i$.
Hence $\iota(s)$ = $\sigma$.
Hence $\iota$ is surjective.
QED
Lemma 6
Let $\mathcal F$ be a presheaf on a toplogical space $X$.
Then $\mathcal F^+_x$ = $\mathcal F_x$ for each $x \in X$.
Proof:Clear.
Lemma 7
Let $\mathcal F$ and $\mathcal G$ be presheaves on a toplogical space $X$.
Let $f:\mathcal F \rightarrow G$ be a morphism.
Let $U$ be an open subset of $X$.
Let $\sigma \in \mathcal F^+(U)$.
Then the map $f^+_U(\sigma):U \rightarrow G^+(U)$ which sends $x \in U$ to $f_x(\sigma(x))$ for each $x \in U$ belongs to $\mathcal G^+(U)$.
Proof:Clear.
Lemma 8
Let $\mathcal F$ and $\mathcal G$ be presheaves on a toplogical space $X$.
Let $f:\mathcal F \rightarrow G$ be a morphism.
Let $\iota_{\mathcal F}:\mathcal F \rightarrow \mathcal F^+$ and $\iota_{\mathcal G}:\mathcal G \rightarrow \mathcal G^+$ be the canonical morphisms.
Then there exists a unique morphism $f^+:\mathcal F^+ \rightarrow \mathcal G^+$ such that
$f^+\iota_{\mathcal F} = \iota_{\mathcal G} f$.
Proof:
There exists the canonical morphism $f_x:\mathcal F_x \rightarrow \mathcal G_x$ for each $x \in X$.
Let $U$ be an open subset of $X$.
We define a map $f^+_U:\mathcal F^+(U) \rightarrow \mathcal G^+(U)$ by sending $\sigma \in \mathcal F^+(U)$ to $f^+_U(\sigma) \in \mathcal G^+(U)$, where $f^+_U(\sigma)$ is defined in Lemma 7.
Clearly this gives a morphism of presheaves $f^+:\mathcal F^+ \rightarrow \mathcal G^+$ and $f^+\iota_{\mathcal F} = \iota_{\mathcal G} f$.
It remains to prove the uniqueness of $f^+$.
Let $\psi:\mathcal F^+ \rightarrow \mathcal G^+$ be a morphism such that
$\psi\iota_{\mathcal F} = \iota_{\mathcal G} f$.
Since $\mathcal F^+_x$ = $\mathcal F_x$ by Lemma 6, $\psi_x$ = $f^+_x$ for each $x \in X$. Since $\mathcal F^+$ and $\mathcal G^+$ are sheaves by Definition 1, $\psi$ = $f^+$.
QED
Proposition
Let $\mathcal F$ be a presheaf on a toplogical space $X$.
Let $\mathcal G$ be a sheaf on a toplogical space $X$.
Let $f:\mathcal F \rightarrow \mathcal G$ be a morphism.
Then there exists a unique morphism $\theta:\mathcal F^+ \rightarrow \mathcal G$ such that
$\theta\iota$ = $f$, where $\iota:\mathcal F \rightarrow \mathcal F^+$ is the canonical morphism.
Proof:
This follows immediately from Lemma 5 and Lemma 8.
Here's a concrete way to go back and forth between the descriptions.
A continuous function $\sigma:U \rightarrow LF$ satisfying $p\circ \sigma = \mathbb 1$ corresponds to the element $(\sigma (x)) \in \prod_{x\in U} F_x$. To show that $(\sigma(x)) \in F^{\#}(U)$, let $a \in U$ and let $U'\subset LF$ be so that $\sigma(a) \in U'$. Since basic open sets of $LF$ are of the form $s[V]$ for $s \in F(V)$, we may assume $U' = s[V]$. By continuity of $\sigma$, there exists a neighborhood $W$ of $a$ so that $\sigma(W) \subset s[V]$. But this implies $\sigma(b)=s_b$ for all $b \in W$. Since this holds for all $a \in U$, $(\sigma(a))$ satisfies $(*)$.
For the other direction, take $(s_x)\in F^{\#}(U)$ and let $\sigma$ be the section of $p$ over $U$ defined by $x \mapsto s_x$. We must verify $\sigma$ is continuous. For any $x \in U$ and any basic open set $t[V]\subset LF$ with $s_x \in t[V]$, $(*)$ implies that there exists $W \subset V$ so that $s_y = t_y$ for all $y \in W$. This shows that $x\in\sigma(W) \subset t[V]$, and hence $\sigma$ is continuous.
To completely nail down the equivalence of the definitions, you'll want to check that these identifications of $\Gamma LF(U)$ and $F^{\#}(U)$ for $U \subset X$ open establish an isomorphism of sheaves (i.e. that the maps $\Gamma LF(U) \rightarrow F^{\#}(U)$ are morphisms and the appropriate diagrams involving restrictions commute).
Best Answer
So, basically the sections of the associated sheaf are functions which are locally sections of the presheaf. Maybe a clear way to see this is the following example: take a smooth manifold $M$ and define a presheaf $\mathscr{F}$ by $\mathscr{F}(U)=\{f:U\to \mathbb{R}:f\:\text{is constant}\}$. If $M$ is not connected, then write $M_1\sqcup M_2$ for instance, both clopen. We can write a function $f_1:M_1\to \mathbb{R}$ by $f_1(x)=1$ and $f_2:M_2\to \mathbb{R}$ by $f_2(x)=2$. Because $M_1\cap M_2=\varnothing$, if $\mathscr{F}$ were a sheaf, then the piecewise function $f:M\to \mathbb{R}$ given by $$f(x)= \begin{cases} f_1(x)&x\in M_1\\ f_2(x)& x\in M_2. \end{cases}$$ would have to be in the sheaf. However, it is not because $f$ is not constant globally, only locally. Once we pass to the sheafification, $\mathscr{F}^+$, our sections on $U\subseteq M$ are $$ \mathscr{F}^+(U)=\{f:U\to \mathbb{R}:(1)\:\&\:(2)\}.$$ $(1)$ is the property that $f(P)\in \mathscr{F}_P$. $\mathscr{F}_P=\mathbb{R}$ so this is fine.
$(2)$ is the property that around each $P\in U$, there exists a neighborhood $V\subseteq U$ with $f|_V(P)=s_P$ for $s\in \mathscr{F}(U)$. That is, $f$ is locally a member of $\mathscr{F}$. That is, $f$ is locally constant.
So, the sections of $\mathscr{F}^+(U)$ are locally constant functions valued in $\mathbb{R}$. Morally, a sheaf is supposed to be an object that locally assigns functions on the space, and the sheafification formalism says that indeed this is the case.