There are two ways a presheaf can fail to be a sheaf.
- It has local sections that should patch together to give a global section, but don't,
- It has non-zero sections which are locally zero.
When dividing the problems into two classes, it is easy to see what sheafifying does. It adds the missing sections from the first problem, and it throws away the extra sections from the second problem.
The latter kind of sections tend to be easier to notice, but are less common. Usually, when a construction or functor must be sheafified, it has local sections that should patch together but don't.
A simple example of a presheaf with this property is the presheaf $F_{p=q}$ of continuous functions on the circle $S^1$ which have the same value at two distinct points $p,q\in S^1$. When I restrict to an open neighborhood of $p$ that doesn't have $q$, the condition on their values goes away. Because the same thing is true for open neighborhoods of $q$ which don't contain $p$, the condition on the functions in this presheaf has no effect on sufficiently small open sets. It follows that this presheaf is locally the same as the sheaf of continuous functions. Therefore, for any function on $S^1$ which has different values on $p$ and $q$, I can restrict it to an open cover where each local section is in $F_{p=q}$, but this function is not in $F_{p=q}$. This is why $F_{p=q}$ is not a sheaf.
When we sheafify, we just add in all these sections, to get the full sheaf of continuous functions. This is clear, because any two sheaves which agree locally are the same (though, I mean that the local sections and local restriction maps agree).
This example really does come up in examples. Consider the map $S^1\rightarrow \infty$, where $\infty$ is the topological space which is $S^1$ with $p$ and $q$ identified. If I pull back the sheaf of functions on $\infty$ in the naive way, the resulting presheaf on $S^1$ is $F_{p=q}$. To get a sheaf, we need to sheafify.
Let's label the three conditions you wrote as 1'), 2'), and 3)'. Combine 1') and 2') by saying for any finite cover by compact subsets... (I hope you follow what I mean). Also, in 3)', you better mean that $\bar U$ is compact, otherwise it doesn't make sense.
So now we're reduced to 1) and 2). Also, let's not talk about sheaves of abelian groups, but sheaves of sets. Sheaves of abelian groups are just abelian group objects in the category of sheaves of sets, so there is no harm to start with the latter. Now, if you want to say you are doing "sheaf theory" you better actually have a Grothendieck topology. Here is what you can do in your situation:
Given your space $X$, define the category $K\left(X\right)$ as the poset of of compact subsets of $X$ and their inclusions. You can define a Grothendieck pre-topology, by saying a covering family is a finite family of jointly surjective inclusions. Being a sheaf for this topology is equivalent to condition 1) (i.e. 1') and 2)'). Let us call this topology $J$.
Consider the category $O_c\left(X\right)$ of open subsets of $X$ which have compact closure. Since $X$ is locally compact, these form a basis for $X$. There is a Grothendieck pre-topology on this category which is the usual one (restricted to this subcategory), except we only allow finite covering families. Sheaves for the associated Grothendieck topology will in general not be sheaves on $X$ in the classical sense, unless $X$ is compact. However, given a sheaf $F$ on $K(X)$, we can define $F(U)$ for a $U$ in $O(X)_c$ by $\varinjlim F(K)$ running over all compacts containing $U,$ but note, this is the same as $F(\bar U),$ since the poset of compact subsets containing $U$ has $\bar U$ as a terminal object. Note, we may also just simply remark that there is a functor $$cl:O_c(X) \to K(X)$$ induced by taking closures and what we have done is defined $cl^*F$. Then, 1) implies that $cl^*F$ is a sheaf for finite open covers as well, since $X$ is locally compact. It would seem that this implies 2) follows automatically, however I haven't checked carefully.
EDIT: This fails in general! Condition 2) is equivalent to for all $F$ sheaves on $K(X)$, $cl_!cl^*F \cong F,$ i.e. the co-unit of the adjunction $$cl_! \dashv cl^\star$$ needs to be an iso on $Sh(K(X))$, which is if and only if $cl^{\star}$ is full and faithful when restricted to $Sh(K(X)).$ Since $cl$ is itself full and faithful, it implies that $cl_*$ is, and hence we get that $F$ must lie in the image of the pullback topos $$Sh(O_c(X)) \times_{Set^{K(X)^{op}}} Sh(K(X))$$ in $Sh(K(X))$. Concretely, this means there is a finer Grothendieck topology $J'$ on $K(X)$ which does the trick. It is possible to write down explicitly, what the covers are, but I will not attempt to do so here. To get an idea of how to do this though, look at how I construct the "compactly generated Grothendieck topology" here: http://arxiv.org/abs/0907.3925
*The following should work with either $J$ or $J'$ *
If all this is right, then all you are talking about is a sheaf on $K(X)$ with the Grothendieck topology I have mentioned. Given such a sheaf $F,$ if you restrict to a compact $C$, $F|C$ defines an ordinary sheaf on $C,$ by extending it to opens the way I described. Since every point $x$ in $X$ has a compact neighborhood, it should follow that sections are determined by their stalks, by reducing to the case of ordinary sheaves on compact spaces.
Now, if $i$ is a closed inclusion, $i:V \to X,$ then $i$ induces a functor $$i^{-1}:K(X) \to K(V),$$ by intersecting with $X.$ This functor preserves covers, so the functor $$i_*:Sh(K(V)) \to Sh(K(X)),$$ defined by $$i_*F(C)=F(C \cap V)$$ for all compact $C$ in $X,$ is well defined. It has a left adjoint $i^{\star},$ which is simply given by procompositing with $i.$
Note: The following probably fails now, with the new topology:
I am not sure how to get $j_!$ for open inclusions, since for open covers this is usually induced by the inclusion functor $O(U) \to O(X),$ but we have no inclusion functor $K(U) \to K(X).$ BUT, we do have an inclusion functor $$K(V) \to K(X);$$ it is right adjoint to the functor $i^{-1}.$ It also preserves covers, so we get an induced functor $$i_!:Sh(K(V)) \to Sh(K(X))$$ by left Kan extension. It is in fact left adjoint to $i^*$ so, from a closed inclusion we get both $i_*$ and $i_!.$
Cech cohomology is automatic, since it makes sense in any topos. This is always true
Best Answer
Let me try and take a stab at this question. I will give not so much as an answer to your question, but more of a rambling collection of remarks. The post turned out to be much longer than I wanted, because as B. Pascal once complained, I did not have the time to write a shorter one. Expect some inconsistencies, a lot of hand-waving, etc.
To put some order in my thoughts, I will try to argue the following four points. Fix a category $\mathcal{A}$, which is the category "of interest". Then:
(1) Functors into $\mathcal{A}$ are interesting.
(2) Functors into $\mathcal{A}$ tell you everything about $\mathcal{A}$.
(3) Part of the perceived asymmetry follows from the fact that in many cases the interesting category $\mathcal{A}$ is equivalent to a presheaf category (or some localization thereof).
(4) There is not really an asymmetry between functors out of and functors into $\mathcal{A}$, but more of a duality.
The first point is easy to settle, as Qiaochu Yuan already gave a class of interesting examples: for a category of interest $\mathcal{A}$ and any category $\mathcal{I}$, a functor $\mathcal{I}\to \mathcal{A}$ is a diagram of shape $\mathcal{I}$ in $\mathcal{A}$, so hell yeah, functors into $\mathcal{A}$ are important. You may think this a rather pedestrian example, but below I give more examples.
To explain (2), instead of working with categories, that is, in the $2$-category of categories, let us go one level down to the category of sets. A set $X$ is determined completely by its elements, that is, maps $\ast\to X$ where $\ast$ is a singleton set (any one will do: for the sake of determinacy, take the singleton set comprised of the element $\emptyset$). Now in a general category, we cannot speak of elements, but we can (and do) speak of generalized elements.
Definition: Let $\mathcal{A}$ be a category and $a$ an object of $\mathcal{A}$. A generalized element of $a$ is a map $x\colon d\to a$. This is also written symbolically as $x\in_{d} a$.
We do not need generalized elements to develop category theory, but personally, I found them useful to build upon the intuition gained from working with more "concrete" categories. A very nice discussion of generalized elements is in Awodey's book on category theory. Now the kicker: Yoneda's lemma tells us that if we know all the generalized elements of an object $a$ then we know everything about $a$, including of course, the arrows out of $a$ (note: and by duality, if we know all the arrows out of $a$ we will know everything about $a$). Two possible objections may be raised:
The original example involves a $2$-category: does not make much of a difference, as my reply to Kevin's comment (see above) still applies. And besides, there are $2$-categorial versions of Yoneda lemma.
Yoneda's lemma works ok, but you need the knowledge of all generalized elements and these range over a potentially proper class of objects: that is true, but in virtually every interesting category one can trim down this proper class to a (small) set, even a singleton set. I will not spell out the proper definitions; they are intimately tied with "smallness" conditions and the adjoint functor theorems (and yes, the category of (small) categories satisfies them).
For (3), let $\mathcal{C}^{\mathcal{B}}$ be a functor category. For size reasons, $\mathcal{B}$ has to be small (note: if you have no scientifick problems with creation ex nihilo, you can always spawn a larger universe by invoking the axiom of universes and sidestep this particular size issue). Since functor categories inherit most of the good properties of the codomain category, you will want $\mathcal{C}$ to be as good behaved as possible (e.g. complete, cocomplete, abelian, symmetric monoidal closed, etc.). It is not true that the structure of $\mathcal{B}$ is irrelevant for the structure of $\mathcal{C}^{\mathcal{B}}$, but it is true that in general, $\mathcal{B}$ only needs a bare minimum of structure. This asymmetry between the domain and the codomain categories is reinforced by the fact that many of the interesting categories $\mathcal{A}$ are equivalent to functor categories, even presheaf categories (or some localization of them). Here are two examples.
Let us consider the category of groups $\operatorname{Grp}$, undoubtebly a category "of interest". There is a category $\operatorname{Th}(\operatorname{Grp})$ that has all finite products such that $\operatorname{Grp}$ is equivalent to the category of product-preserving functors $\operatorname{Th}(\operatorname{Grp})\to \operatorname{Set}$ where $\operatorname{Set}$ is the category of sets. This equivalence can be generalized to a very large class of categories of "algebraic flavor" and even some that at first sight do not bear the least resemblance to "algebraic categories". A few extra remarks about this example. First, the category $\operatorname{Th}(\operatorname{Grp})$ is a category constructed to make the equivalence work (it is the free category with products on a group object), in other words, it does not exactly fall within the class of categories "of interest". It's a similar to the example of diagram categories in (1), where the domain, a free category on a graph, is just a categorial construction to make the identification of diagrams with functors, not an interesting category by itself. Second, by replacing $\operatorname{Set}$ by another category $\mathcal{B}$ (with at least finite products), you can now speak of groups in $\mathcal{B}$. This gives another class of examples where functors into a category are interesting.
If $(\mathcal{C}, \mathcal{J})$ is a site (a category with a Grothendieck topology), by first taking the category of presheaves and then a suitable localization, one obtains the category of sheaves. This produces a host of geometric categories, like manifolds and schemes.
Much like in example 1, the interest is not so much on $(\mathcal{C}, \mathcal{J})$ and even less in the codomain, which is usually the category of sets (other categories for codomain also work, but the categorial requirements for everything to work smoothly are fairly strong), but in the (pre)sheaf category.
For my last point (4), a lot could be said, but I will just point you to two articles by F. W. Lawvere in the TAC Reprints, "Metric Spaces, Generalized Logic and Closed Categories" and "Taking categories seriously" (google for them, they are available online). In them, Lawvere makes several remarks about the duality between spaces and algebras of functions which are directly relevant to your question. To show that there is not so much an asymmetry but a duality between functors into and out of, let me give you two examples.
In your (that is, the OP) last post, you speak about stacks. A stack on a category $\mathcal{A}$ can be defined as a functor with values in $\mathcal{A}$ satisfying some conditions -- this is the fibered category approach to stacks. But a stack can also be defined as weak $2$-functor with values in the $2$-category of categories (satisfying some extra conditions). For reasons that I will not explain, the first approach is better, but nevertheless the point should be clear. There are actually many examples of this "duality", that identifies some category of functors into $\mathcal{A}$ with some category of functors out of $\mathcal{A}$.
Let me end up with an example from physics that further illustrates this duality. Quantum field theories are notoriously hard objects to define (let alone study). Several years ago, V. Turaev defined the notion of a Homotopy Quantum Field Theory, HQFT for short (check his papers in the arxiv if you are interested) which is a very simple, "toy" example of a QFT. If $X$ is a topological space, we can define a category that has for objects manifolds $M$ equipped with a homotopy class of maps $g\colon M\to X$ and a morphism $(M, g)\to (N, h)$ is a cobordism $W\colon M\to N$ with a homotppy class of maps into $X$ extending $g$ and $h$. An HQFT is a monoidal functor from this category into another monoidal category (usually, the category of finite-dimensional complex linear spaces). I am omitting lots of details, but the gist is that an HQFT gives us invariants of manifolds $M$ by mapping $M$ into some fixed background space $X$. But we can turn things around, for the category of $X$-HQFT's is a (functorial) invariant of the homotopy type of $X$, an invariant cooked up by mapping manifolds into $X$.
Hope it helps, regards, G. Rodrigues