Why is the standard definition of cocycle the one that _always_ comes up?? This question might not have a good answer. It was something that occurred to me yesterday when I found myself in a pub, needing to do an explicit calculation with 2-cocycles but with no references handy (!).
Review of group cohomology.
Let $G$ be a group acting (on the left) on an abelian group $M$. Then $H^0(G,M)=M^G=Hom_{\mathbf{Z}[G]}(\mathbf{Z},M)$ and hence $$H^i(G,M)=Ext^i_{\mathbf{Z}[G]}(\mathbf{Z},M).$$ Now $Ext$s can be computed using a projective resolution of the first variable, so we're going to get a formula for group cohomology in terms of "cocycles over coboundaries" if we write down a projective resolution of $\mathbf{Z}$ as a $\mathbf{Z}[G]$-module.
There's a very natural projective resolution of $\mathbf{Z}$: let $P_i=\mathbf{Z}[G^{i+1}]$ for $i\geq0$ (with $G$ acting via left multiplication on $G^{i+1}$) and let
$d:P_i\to P_{i-1}$ be the map that so often shows up in this sort of thing: $$d(g_0,\ldots,g_i)=\sum_{0\leq j\leq i}(-1)^j(g_0,\ldots,\widehat{g_j},\ldots,g_i).$$
Check: this is indeed a resolution of $\mathbf{Z}$. So now we have a "formula" for group cohomology.
But it's not the usual formula because I need to do one more trick yet. Currently, the formula looks something like this: the $i$th cohomology group is $G$-equivariant maps $G^{i+1}\to M$ which are killed by $d$ (that is, which satisfy some axiom involving an alternating sum), modulo the image under $d$ of the $G$-equivariant maps $G^i\to M$.
The standard way to proceed from this point.
The "formula" for group cohomology derived above is essentially thought-free (which was why I could get this far in a noisy pub with no sources). But we want something more useful and it was at this point I got stuck. I remembered the nature of the trick: instead of a $G$-equivariant map $G^{i+1}\to M$ we simply "dropped one of the variables", and considered arbitrary maps of sets $G^i\to M$. So we need to give a dictionary between the set-theoretic maps $G^i\to M$ and the $G$-equivariant maps $G^{i+1}\to M$. I could see several choices. For example, given $f:G^{i+1}\to M$ I could define $c:G^i\to M$ by $c(g_1,g_2,\ldots,g_i)=f(1,g_1,g_2,\ldots,g_i)$, or $f(g_1,g_2,\ldots,g_i,1)$, or pretty much anything else of this nature. The point is that given $c$ there's a unique $G$-equivariant $f$ giving rise to it. Which choice of dictionary do we use. Each one will give a definition of group cohomology as "cocycles" over "coboundaries". But which one will give the "usual" definition? Well---in some sense, who cares! But at some point maybe someone somewhere made a decision as to what the convention from moving from $f$ to $c$ was, and now we all stick with it. The standard decision was the rather clunky
$$c(g_1,g_2,\ldots,g_i)=f(1,g_1,g_1g_2,g_1g_2g_3,\ldots,g_1g_2g_3\ldots g_i).$$
Why is the standard definition ubiquitous?
Actually though, I bet that no-one really made that decision. I bet that the notion of a 1-cocycle and a 2-cocycle preceded general homological nonsense, and the dictionary between $f$s and $c$s was worked out so that it agreed with the definitions which were already standard in low degree.
But this got me thinking: if, as it seems to me, there is no "natural" way of moving from $f$s to $c$s, then why do the 1-cocycles and 2-cocycles that naturally show up in mathematics all satisfy the same axioms??. Why doesn't someone do a calculation, and end up with $c:G^2\to M$ satisfying some random axiom which happens not to be the "standard" 2-cocycle axiom but which is an axiom which, under a non-standard association of $c$s with $f$s, becomes the canonical cocycle axiom for $f$ that we derived without moving our brains? Does this ever occur in mathematics? I don't think I've ever seen a single example.
In some sense it's even a surprise to me that there is a uniform choice which specialises to the standard choices in degrees 1 and 2.
Examples of cocycles in group theory.
1) Imagine you have a 2-dimensional upper-triangular representation of a group $G$, so it sends $g$ to the $2\times 2$ matrix $(\chi_1(g),c(g);0,\chi_2(g))$. Here $\chi_1$ and $\chi_2$ are group homomorphisms. What is $c$? Well, bash it out and see that $c$ is precisely a 1-cocycle in the sense that everyone means when they're talking about 1-cocycles. So we must have used the dictionary $c(g)=f(1,g)$ when moving between $c$s and $f$s above. Why didn't we use $c(g)=f(g,1)$? 
2) Imagine you're trying to construct the boundary map $H^0(C)\to H^1(A)$. Follow your nose. Claim: your nose-following will lead you to the standard cocycle representing the cohomology class. Why did it not lead to a non-standard notion? Why do the notions (1) and (2) agree? 
3) Imagine we have a group hom $G\to P/M$ where $M$ is an abelian normal subgroup of the group $P$. Can we lift it to a group hom $G\to P$? Well, let's take a random set-theoretic lifting $L:G\to P$. What is the "error"? A completely natural thing to write down is the map $(g,h)\mapsto L(g)L(h)L(gh)^{-1}$ because this will be $M$-valued. This is a 2-cocycle in the standard sense of the word. Why did the natural thing to do come out to be the standard notion of a 2-cocycle? Aah, you say: there are other natural things that one can try. For example we could have sent $(g,h)$ to $L(gh)^{-1}L(g)L(h)$. For a start this "looks slightly less natural to me" (why does it look less natural? That's somehow the question!). Secondly, this is really just applying a canonical involution to everything: we're inverting on $G$ and inverting on $M$, which is something that we'll always be able to do. It certainly does not correspond to the "more natural" dictionary that I confess I tried down the pub, namely $c(g,h)=f(1,g,h)$, which gave much messier answers. Why does my "obvious" choice of dictionary lead me to 20 minutes of wasted calculations? Why is it "wrong"?
The real question.
Has anyone ever found themselves in a situation where a natural cocycle-like construction is staring them in the face, and they make the construction, and find themselves with a non-standard cocycle? That is, a cocycle which will induce an element of a cohomology group but only after a non-standard dictionary is applied to move from $f$s to $c$s?
Edit: Here is what I hope is a clarification of the question. To turn the cocycles from $G$-equivariant maps $G^{i+1}\to M$ to maps of sets $G^i\to M$ we need to choose a transversal for the action of $G$ on $G^{i+1}$ and identify this with $G^i$. There seems to be one, and only one, way to do this that gives rise to the "standard" definition of n-cocycle that I (possibly incorrectly) percieve to be ubiquitous in mathematics. I call this "the clunky way" because the map seems odd to me. Why is it so clunky? And why, whenever a 2-cocycle falls out of the sky, does it always seem to satisfy the axioms induced by this clunky method? Why aren't there people popping up in this thread saying "well here's a completely natural "cocycle" $c(g,h)$ coming from theory X that I study, and it doesn't satisfy the usual cocycle axioms, we have to modify it to be $c'(g,h)=c(g,gh)$ before it does, and this boils down to the fact that in theory X we would have been better off if history had chosen a non-clunky identification?"
 A: The "trick" you are referring to, of replacing $G$-linear morphisms $f:\mathbb ZG^{i+1}\to M$ by functions of sets $c:G^i\to M$, is not a trick: it is just the observation that $G$-linearity implies that, since $\mathbb{Z} G^{i+1}=\mathbb ZG\otimes\mathbb Z{G^i}$ as a left $G$-module, there is a natural isomorphism $$\hom_G(\mathbb ZG^{i+1},M)=\hom_G(\mathbb ZG\otimes\mathbb Z{G^i},M)=\hom_{\mathbb Z}(\mathbb Z G^i,M)$$ by standard properties of the tensor product. Now $\mathbb ZG^i$ is free abelian on the set $G^i$, so there is a natural isomorphism $$\hom_{\mathbb Z}(\mathbb Z G^i,M)\cong\hom_{\mathrm{Set}}(G^i,M).$$ The composition of theses maps gives the identification between $f$'s and $c$'s. You can write them down explicitly, and compute their inverses: in both cases a one on the left" is involved and this is simply a reflection that we are using left modules.  

On clunkiness
The form you call 'clunky' has the following origin. The usual bar resolution for a group $G$ has in degree $n$ the $\mathbb ZG$-module $P_n=\mathbb ZG^{n+1}$. We can think of it as a free left $G$-module generated by the set of symbols $[g_1,\dots,g_n]$. The differential of the complex is then left $G$-linear and for example, $$d[g_1,g_2,g_3] = g_1[g_2,g_3]-[g_1g_2,g_3]+[g_1,g_2g_3]-[g_1,g_2],$$
and this plainly involves the multiplication of the group. This is the so-called inhomogeneous description.
We can also think of $P_n$ in another, homogenous, way: define $$(g_0,\dots,g_n)=g_0[g_0^{-1}g_1,\dots,g_0^{-1}g_n].$$ The set of all such symbols is now a basis of $P_n$ over $\mathbb Z$, with the left action of $G$ given by $$g\cdot(g_0,\dots,g_n)=(gg_0,\dots,gg_n),$$ which is a bit more annoying that before. The good thing is that now the boundary is given by $$d(g_0,\dots,g_n)=\sum_{i=0}^n(-1)^i(g_0,\dots,\hat g_i,\dots g_n),$$ which does not involve the product in $G$ at all and, in fact, is precisely the same formula as the one giving the boundary in a simplicial complex.
The clunkiness you refer to is nothing but the translation formula between these two descriptions of the complex, as $$[g_1,\dots,g_n]=(1,g_1,g_1g_2,g_1g_2g_3,\dots,g_1\cdots g_n).$$
The original description of the complex is the inhomogeneous one, because it was found topologically. The two can be seen as a description of the classifying space for $G$ either by labelling paths in the category by the objects you go through, or by the arrows you follow, or something...
A: Group cohomology of a group $G$ looks at homotopy classes of maps $\mathbf{B}G \to \mathbf{B}^n \mathbb{Z}$ from an oo-groupoid $\mathbf{B}G$ that is a 
delooping  of $G$ to, in the simplest case, some Eilenberg-MacLane oo-groupoid $\mathbf{B}^n \mathbb{Z}$.
So  $\mathbf{B}G$ here, like everything else in the game, is specified only up to equivalence. There are many ways to realize this.
One is to realize it as the action groupoid $ {*} // G$ of $G$ acting trivially on the point. That gives the second formula that you mention.
But we can take bigger models of the point. One popular one is $\mathbf{E}G = G//G$, the action groupoid of $G$ acting on itself. This is an equivalent model of the point. So the cohomology of $\mathbf{E}G//G$ also computes the group cohomology of $G$. But in components maps out of $\mathbf{E}G/G$ are just maps out of copies of $G$, with one more copy than we had before, that respect the $G$ action. This is the first formual you have.
There are infinitely more, and infinitely more complicated ones than you listed. One for each connected oo-groupoid who looping gives $G$, up to equivalence.
Generally, cohomology is something that lives in a homotopical context, where everything only depends on everything else up to weak equivalence. That's why there are, generally, so many different models for cohomologies. Not just group cohomology.
Some details on this are at nLab: group cohomology 
A: The difference between f(1,g) and f(g,1) is generally an issue of whether mathematicians give preference to "domains" or "ranges" of maps.
Here is one way that you could think of this.  I can write EG for a category whose objects are objects are elements of G, and where each pair of objects has a unique map between them.  This category has an action of G on it, and you can ask about G-equivariant functors from this category to another category what has a G-action on it.
To define such a functor on the level of objects it suffices to define F(1), where 1 is the unit; equivariance forces us to define F(g) = g F(1).  On the level of morphisms, however, we have to make a choice.  The unique morphism g→h becomes a morphism F(g)→F(h), and to make such maps compatible with the G-action it suffices to make one of the following sets of choices:


*

*We could define maps fh:F(1)→F(h) for all h, and get all the other maps as g fh:F(g)→F(gh).  To be a functor, we need this to satisfy the cocycle condition fgh = (g fh) fg.

*We could define maps dh:F(h)→F(1) for all h, and get all the other maps as g dh:F(gh)→F(g).  To be a functor, we need this to satisfy the cocycle condition dgh = dg (g dh).


In group cohomology, H1(G,M) classifies splittings in the semidirect product of G with M, and the cocycle condition we get comes from our convention of writing this group as pairs (m,g) (which is in the same order as the exact sequence it fits into) and not (g,m).  Similarly for H2(G,M).
I would say that I've hit nonstandard cocycle definitions several times because I've been too lazy to come up with sensible conventions about when I'm thinking about domains and ranges or trying to sweep it under the rug, especially when dealing with Hopf algebroids and cohomological calculations there.
I don't have a good answer for higher cocycle conditions other than saying that writing 2-cochains using f(g,1,h) is somehow more unusual than either of the other 2 choices because it's somehow derived from focusing on the "middle" object in a double composite of maps.
A: Let me just beat a dead horse a little.  As Mariano and DC have mentioned (and you mentioned in your original question), there are multiple equivalent ways to compute group cohomology, or Ext in general, because you can pick any resolution you want.  However, if you want a systematic resolution that always works then you need to work harder.
The resolution that you wrote down in terms of G acting on Gi+1 works perfectly well but it depends on a special fact about the group algebra: it's a Hopf algebra, and as a result it has a sensible way to act on a tensor product of copies of itself.  Not all rings have such properties.
The "clunky" definition works for more general rings.  Given any ring R and an R-module M, there is always an exact sequence
$$\cdots \to R \otimes R \otimes M \to R \otimes M \to M \to 0$$
called the bar resolution of R over M, and it arises from a simplicial construction via some natural adjoint-functor considerations - namely, there is an adjoint pair of the forgetful functor from R-modules to ℤ-modules and its left adjoint, the free R-module functor, and together they construct a simplicial object which you convert into a chain complex etc etc.  If you specialize this construction to R = ℤ[G], you recover the "clunky" resolution and its specific formula for the boundary and coboundary operators.  So somehow the definition that is less intuitive comes from forgetting that we're working in a grouplike context and being systematic there.
Because I can't help myself, I should mention that the above bar resolution is only a free resolution if R and M are free over ℤ (or if you replace ℤ and tensor with some ground ring over which they are both free).  If not, then in order to get a canonical resolution you have to dip all the way down to the forgetful functor from R-modules to sets and its adjoint, the free R-module on a set, and this gives you an absolutely nightmarish but always-valid resolution that is even less intuitive.
A: Late to the party as usual, but: the goal of this answer is to convince you that the standard convention for $2$-cocycles is so natural that you should consider it perverse to consider any other convention, modulo "applying a canonical involution to everything," as you say. To keep things simple let's only deal with trivial action on coefficients. The motivating question is the following:

What does it mean for a group $G$ to act on a category $C$?

For starters we should attach to each element of $G$ a functor $F(g) : C \to C$. Next we could require that $F(g) \circ F(h) = F(gh)$, but we should really weaken equalities of functors to natural isomorphisms whenever possible. Hence we should attach to each pair of elements of $G$ a natural isomorphism
$$\eta(g, h) : F(g) \circ F(h) \to F(gh).$$
This is the point at which we pick a convention for how we're going to represent $2$-cocycles. Instead of talking about $\eta(g, h)$ we could talk about its inverse; which we choose corresponds to whether we prefer to talk about lax monoidal or oplax monoidal functors, since what we're going to end up writing down is a lax monoidal resp. an oplax monoidal functor from $G$ (regarded as a discrete monoidal category) to $\text{Aut}(C)$ (regarded as a monoidal category under composition). 
In any case, let's stick to the above choice (the lax one). Then the isomorphisms $\eta(g, h)$ should satisfy some coherence conditions, the important one being the "associativity" condition that the two obvious ways of going from $F(g_1) \circ F(g_2) \circ F(g_3)$ to $F(g_1 g_2 g_3)$ should agree.
Now let's assume that in addition all of the functors $F(g)$ are the identity functor $\text{id}_C : C \to C$. Then the only remaining data in a group action is a collection of natural automorphisms
$$\eta(g, h) : \text{id}_C \to \text{id}_C$$
of the identity functor. For any category $C$, the natural automorphisms of the identity functor naturally form an abelian (by the Eckmann-Hilton argument) group which here I'll call its center $Z(C)$ (but this notation is also used for the commutative monoid of natural endomorphisms of the identity). So we get a function
$$\eta : G \times G \to Z(C).$$
The important coherence condition I mentioned above now reduces (again by the Eckmann-Hilton argument) to the condition that for any $g_1, g_2, g_3 \in G$ we have
$$\eta(g_1, g_2) \eta(g_1 g_2, g_3) = \eta(g_2, g_3) \eta(g_1, g_2 g_3)$$
which is precisely the standard cocycle condition. (Coboundaries come in when you ask what it means for two group actions to be equivalent; I'm going to ignore this.) 
The only reason this condition, which recall is in general just the statement that the two obvious ways of going from $F(g_1) \circ F(g_2) \circ F(g_3)$ to $F(g_1 g_2 g_3)$ should agree, could ever have looked anything other than completely natural is that it's a degenerate special case where the sources and targets of the various maps involved have been obscured because they are identical. In particular, of course I could have instead chosen to think about the natural isomorphisms
$$\eta(g, g^{-1} h) : F(g) \circ F(g^{-1} h) \to F(h)$$
(which corresponds to your $f(1, g, h)$), but now


*

*it's no longer at all obvious how to state the associativity condition succinctly, and

*this requires that I make explicit use of the fact that $G$ is a group.


The discussion up til now in fact gives a perfectly reasonable definition for what it means for a monoid to act on a category. (If I want to weaken "natural isomorphism" to "natural transformation," though, I get two genuinely different possibilities depending on whether I pick lax or oplax monoidal functors.) 
Reflecting on associativity suggests that, for a more "unbiased" point of view, we should consider families of natural isomorphisms
$$\eta(g_1, g_2, \dots g_n) : F(g_1) \circ F(g_2) \circ \dots \circ F(g_n) \to F(g_1 g_2 \dots g_n)$$
and then impose a "generalized associativity" condition that every way of composing them to get a natural isomorphism with the same source and target as $\eta(g_1, g_2, \dots g_n)$ should give $\eta(g_1, g_2, \dots g_n)$. Another way to say this is that the cocycle condition (in the $F(g) = \text{id}_C$ special case, at least) should really be written
$$\eta(g_1, g_2, g_3) = \eta(g_1, g_2) \eta(g_1 g_2, g_3) = \eta(g_2, g_3) \eta(g_1, g_2 g_3).$$
This is in the same way that we can consider a monoid operation to be a family $m(g_1, g_2,  \dots g_n) = g_1 g_2 \dots g_n$ of operations satisfying a generalized associativity condition, and in particular satisfying
$$m(g_1, g_2, g_3) = m(m(g_1, g_2), g_3) = m(g_1, m(g_2, g_3)).$$
Namely, by "associativity" we usually mean that the middle expression equals the right, but really the reason that the middle expression equals the right is that they both equal the left. 
A: Hey Kevin - 
You write: "The standard decision was the rather clunky
$$c(g_1,g_2, \cdots g_i)=f(1,g_1,g_1g_2,g_1g_2g_3,\cdots, g_1g_2g_3\cdots g_i)$$
A topologist thinks of group cocycles as operating on "simplices labeled by elements of a group". Since orderings matter, we have "the" standard $n$-simplex, which comes with a labeling of its vertices by the numbers $0,1,2,\cdots,n$. The way the group comes in is that each edge of the simplex from vertex $i-1$ to vertex $i$ is labeled by an element of the group, i.e. $g_i$ is the label going from vertex $i-1$ to vertex $i$. This is because the simplex is "really" part of a $K(G,1)$ with one vertex, and one edge for each element of the group, one triangle for each element of the multiplication table of the group (i.e. for each expression $g \times h = gh$) and so on. These $g_i$ are the terms on the LHS of your equation.
The universal cover of this $K(G,1)$ is a contractible simplicial complex whose vertices are now labeled by elements of the group (because the group "is" the deck group of the cover, and acts simply transitively on vertices). Your original simplex has a unique lift to the cover taking the vertex $0$ to some specific vertex, which might as well be the one labeled by the identity. The vertex $i$ then lifts to the composition $g_1g_2\cdots g_i$. So the terms on the RHS of your equation are the labels on the vertices.
I am really only repeating what Mariano wrote in very slightly more topological language; you didn't seem happy with his answer, so maybe you won't be happy with mine either.
Best,
DC
