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.
Best Answer
As said in the comments, your description of $\mathcal{F}^a$ is not completely correct. First, it should be the disjoint union instead of the direct sum of the $\mathcal{F}_x$. Then all sections of $\pi$ are not allowed, only those that satisfy this condition : $$ (1) \quad \forall x\in U, \exists V\ni x \text{ a neighborhood of $x$ in $U$ and }t\in\mathcal{F}(V) \text{ such that } \forall y\in V, s(y)=t_y $$ Otherwise you have too many sections, for example, if $\mathcal{F}=\mathcal{C}$ is the sheaf of continuous function on a space $X$, with your definition a section in $\mathcal{F}^a$ would consists of the choice of a germ of continuous function at every point, without conditions that these germs glue (and they might define a function which is not continuous).
Hence the good definition is the following : $$\mathcal{F}^a(U)=\{s:U\rightarrow\coprod_{x\in U}\mathcal{F}_x | s \text{ is a section of $\pi$ and satisfies condition $(1)$}\}$$
Now it is easy to see that $\mathcal{F}^a_x=\mathcal{F}_x$. Indeed, if $s_x$ is a germ of a section in $\mathcal{F}^a_x$, then you can find a representative $(U,s)$ where $s\in\mathcal{F}^a(U)$. Now by condition $(1)$, there exists $t\in\mathcal{F}(V)$ such that $\forall y\in V, s(y)=t_y$. But this implies that $(U,s)$ and $(V,t)$ define the same germ. So $s\in\mathcal{F}_x$.
To be perfectly rigorous, check that what I just described is a well-defined map $\mathcal{F}^a_x\rightarrow\mathcal{F}_x$ which is the inverse of the obvious map $\mathcal{F}_x\rightarrow\mathcal{F}^a_x$.