A very optional complement
For the sake of completeness and for the record, I'd like to analyze the nature of the functor (from presheaves to sheaves) $\mathcal F\to \mathcal F^+$ on an example and show that it is subtler than one might think.
Let $X$ be a topological space, $\mathcal C$ be the sheaf of real-valued continuous functions on $X$ and $\mathcal Disc $ the sheaf of all real-valued functions, maybe discontinuous: $\mathcal Disc (U)=\mathbb R^U$.
We have a morphism of sheaves $\mathcal C^+\to \mathcal Disc$ given by
$\mathcal C^+(U)\to \mathcal Disc ( U): s\mapsto \tilde s \text { where} \;\tilde s(x)=(s(x))(x)$
The strange but logical notation $(s(x))(x)$ means that $s(x)=f_x\in \mathcal C_x$ is the germ at $x$ of some continuous function $f$ defined near $x$ and that you then take the value $f(x)=f_x(x)$ of that function at $x$ and obtain the real number $\tilde s(x)$.
It is then true that for every $U\subset X$ the map $\mathcal C^+(U)\to \mathcal Disc ( U)$ is surjective [take germs of constant functions in the domain], so that a fortiori the morphism of sheaves $\mathcal C^+\to \mathcal Disc $ is surjective.
But the morphism of sheaves $\mathcal C^+\to \mathcal Disc $ is not injective, because the morphism $\mathcal C(X)\to \mathcal Disc ( X)$ (for example) is not injective.
Indeed choose $x_0\in X$ and take for $s\in \mathcal C^+(X)$ the collection of germs $s(x)=0_x$ for $x\neq x_0$ and $s(x_0)=g_{x_0}$ where $g$ is a non-zero continuous function (defined near $x_0$) satisfying $g(x_0)=0$.
Then $\tilde s=0\in \mathcal Disc ( X)$ although $s\neq 0\in \mathcal C^+(X)$
If you start with the sheaf $\mathcal C_b$ of locally bounded continuous functions on $X$, you will find $\mathcal C_b^+=\mathcal C^+$ and the same analysis applies .
Conclusion
You may forget all the details above and just remember that $\mathcal C^+$ is not the sheaf $\mathcal Disc$ of arbitrary [and possibly very discontinuous] functions on $X$, but an even much more horrible sheaf !