Short version: The presheaf cokernel of a morphism of sheaves $\mathcal{F'} \rightarrow \mathcal{F}$ is a sheaf if $F'$ is acyclic. A common type of acyclic sheaf are flasque/flabby sheaves, sheaves where all restriction maps are surjective.
Assume we have an exact sequence $0 \rightarrow\mathcal{F'} \rightarrow \mathcal{F} \rightarrow \mathcal{F''} \rightarrow 0.$ Consider the forgetful functor from the category of sheaves to the category of presheaves. This functor is left exact, so we now have a left-exact sequence $0 \rightarrow F' \rightarrow F \rightarrow F''.$ ($F$ is the presheaf corresponding to $\mathcal{F'}$; I'm not aware of the convention regarding this, so this is probably not good notation.)
The derived functor of the forgetful functor gives us a long exact sequence $0\rightarrow F' \rightarrow F \rightarrow F'' \rightarrow H^1F' \rightarrow H^1F\rightarrow H^1F''\rightarrow \cdots$ (once again, I do not know the standard notation for what I call $H^1F$; I'm more used to working with the derived functor of the global sections functor than the forgetful functor.)
So the left-exact sequence of presheaves will be exact (in which case $F''$, a sheaf, corresponds to the cokernel) if $H^1F'$ is $0$. Then the sheaf $F'$ is called acyclic; an important class of acyclic sheaves are flasque (also called flabby) sheaves, which are sheaves where all restriction maps are surjective.
Image is just kernel of cokernel, so if the cokernel is a sheaf, so is the image.
EDIT: Sorry, some of my if and only ifs were incorrect. They should be fixed now.
That categorical definition is for pre-sheaves, the topological definition is for sheaves.
In topological pre-sheaves, a map is surjective if it is epimorphic for each open set $U$ in $X$.
In topological sheaves, however, we instead have to "sheaf-ify" the definition, and we say that the map is "surjective" if the sheaf-ification of the cokernel map is zero.
Basically, in both cases, you have two categories, $\mathcal{Sh}$ and $\mathcal{PSh}$, and in $\mathcal{PSh}$, the "surjective" maps are the ones that are epimorphisms on each $U$, but in the $\mathcal{Sh}$ catageory, you have a more complicated definition of "surjective" (or "epimorphism.")
Consider, instead, two categories, $\mathcal{Ab}$ the category of abelian groups, and $\mathcal{AbTF}$, the full subcategory of "torsion-free" abelian groups - that is, the abelian groups, $A$, where for any $n\in\mathbb Z$ and $a\in A$, $na=0$ iff $n=0$ or $a=0$.
There is the natural inclusion functor $\mathcal{AbTF}\to\mathcal{Ab}$ and a natural adjoint sending $A\to A/N(A)$ where $N(A)$ is the subgroup of nilpotent elements of $A$.
But in $\mathcal{AbTF}$, the "epimorphisms" are not the ones with cokernel (in $\mathcal{Ab}$) $0$, they are the ones with cokerkels which are nilpotent. So, for example, in $\mathcal{Ab}$, the morphism $\mathbb Z\to\mathbb Z$ sending $x\to 2x$ is not an epimorphism, that same map, when considered as a map in $\mathcal{AbTF}$, is an epimorphism.
So consider the "sheafification" functor $\mathcal{PSh}\to \mathcal{Sh}$ to be much like the functor $\mathcal{Ab}\to\mathcal{AbTF}$.
(I believe, but don't quote me, that $f:A\to B$ in $\mathcal{AbTF}$ is an epimorphism if and only if $f\otimes \mathbb Q:A\otimes \mathbb Q\to B\otimes\mathbb Q$ is an epimorphism in $\mathcal{Ab}$.)
Best Answer
For the first question: yes, the morphism of presheaves $i$ is injective, as its kernel is zero.
For the second question: I believe that "natural" here means canonical, since it comes out of the universal property of sheafification. I realise that this is confusing since "natural" has a specific meaning regarding compatibility with functors in category theory.