Directly calculating $H^1_{\mathrm{sing}}(S^1;\mathbb Z)$ While playing around with DeRham and singular cohomology, I decided to try and find out what an element $[\omega]\in H^1_{\mathrm{sing}}(S^1;\mathbb Z)$ would actually look like. The closure condition on $\omega\in C_{\mathrm{sing}}^1(S^1;\mathbb Z)$ means that $\langle\omega,\gamma\rangle\in\mathbb Z$ is independent of the parametrization of the path $\gamma:[0,1]\to S^1$ (up to orientation) and that $\langle\omega,\gamma_1\cdot\gamma_2\rangle=
\langle\omega,\gamma_1\rangle+\langle\omega,\gamma_2\rangle$ where $\gamma_1\cdot\gamma_2$ denotes concatenation of paths. I'd like to have $\omega$ behave like the winding number, and give $\langle\omega,\gamma_0\rangle=1$ for $\gamma_0(t)=e^{2\pi i t}$ the standard parametrization of $S^1$, but I don't see how this can be done without having $\langle\omega,\gamma\rangle\notin\mathbb Z$ for some $\gamma$. So if a generator $\omega$ doesn't behave like a winding number, what does it look like?
 A: The idea of winding numbers is good, but the issue is that there is not a good notion of what a winding number is for something other than a cycle. How often does the arc from $1$ to $i$ wind around the circle?
Apologies for the long answer that follows; this is because it tries to do two things: tell you what a generator actually looks like, and tell you how to better capture your idea of winding numbers.

What one usually does is the following. This tells you how to write down generators of cohomology, but not how to write down geometric generators.
We have the map $d: C_1(S^1) \to C_0(S^1)$. Its image, $B_0(S^1)$, is free abelian (as any subgroup of a free abelian group is free abelian). Therefore, there is some isomorphism $$C_1(S^1) \cong Z_1(S^1) \oplus B_0(S^1)$$ so that the map $d: C_1(S^1) \to B_0(S^1)$ is simply projection to the second factor. This is the first painful step for those of us who like geometric representatives, because this isomorphism is highly non-explicit (you are choosing, for each generator of $B_0(S^1)$, a corresponding chain whose boundary is that generator).
Now we have a further surjective homomorphism $Z_1(S^1) \to H_1(S^1) \cong \Bbb Z$. Again, because the codomain is free abelian, we may write $Z_1(S^1) \cong H_1(S^1) \oplus B_2(S^1)$. 
Thus we have an isomorphism $$\varphi: C_1(S^1) \cong B_2(S^1) \oplus H_1(S^1) \oplus B_0(S^1),$$ respecting all of the differentials we care about. This is horribly non-explicit, but given this, we may construct the desired cocycle $f_\varphi$ by $$f_\varphi(\sigma) = \pi_2 \varphi(\sigma),$$ where this notation means we follow the above isomorphism and then project to the $H_1(S^1)$ factor. 

Here is my favorite notion of geometric representative of an integral cohomology class. A slightly more general version of this captures all cohomology classes which arise from the $\text{Hom}(H_n(M;\Bbb Z), \Bbb Z)$ factor in the universal coefficient theorem. The other factor would arise instead as a linking number construction and require some unpleasant choices.
As I mentioned in the comments, these should be defined on a subcomplex $C$ of geometric nature, so that the inclusion $C \to C_*(M;\Bbb Z)$ is an isomorphism on homology. 
To capture the idea of winding number, we'll do the following: pick a point $x \in S^1$ arbitrarily. Let the subcomplex $C^x_* \subset C_*(M;\Bbb Z)$ consist of those chains $\sigma: \Delta^k \to M$ so that $\sigma$ is a smooth map, and so that both $\sigma$ and $\partial \sigma$ are transverse to $x$. This means that if $k < \dim M$, the map $\sigma: \Delta^k \to M$ should not have $x$ in its image (and similarly for $\partial \sigma$ when $k = \dim M$). If $k = \dim M$, the map $\sigma: \Delta^n \to M$ should have $\partial \sigma$ disjoint from $x$, and for every $p \in \Delta^n$ with $\sigma(p) = x$, we should have $d\sigma_p: \Bbb R^n \to T_x M$ surjective. Similarly for $k > \dim M$.
First, this complex $C^x_* \subset C_*(M;\Bbb Z)$ is quasi-isomorphic to the whole complex. This is essentially because of the transversality theorem: given any smooth manifold $N$ and map $f: N \to M$, we may modify $f$ by a homotopy to be transverse to $x$; and if $f$ is already transverse to $x$ along some closed subset of $N$, we may keep the homotopy fixed along that subset. This allows us to take cycles in $M$ and obtain homologous cycles lying in $C^x$; similarly if $v \in C^x_*$ is a cycle which is a boundary in $C_*(M;\Bbb Z)$, we may modify the bounding chain by a homotopy, fixed on the boundary, so that in fact $v$ is a boundary in $C^x_*$.
Now if $M$ is oriented, I can define a cocycle $\deg_x: C^x_n \to \Bbb Z$, given by $$\deg_x(\sigma) = \#\left(p \in \Delta^n \mid \sigma(p) = x\right),$$ where this sum is weighted by signs: if $d\sigma_p: \Bbb R^n \to T_x M$ is an oriented isomorphism, count $p$ positively, and if not, count it negatively. 
I claim that $\deg_x$ defines a class in $H_n(\text{Hom}(C^x, \Bbb Z)) = H^n(M;\Bbb Z)$, and in fact, this is the usual "fundamental class" of $M$. The picture is that you restrict to those $n$-chains whose boundaries do not cross over $x$, and then $\deg_x(\sigma)$ just counts (with sign) the number of points in the cycle that actually do cross over $x$. This is how one makes sense of the winding number when there is boundary flying around: pick a basepoint to check how often we "wind around at", and make sure the boundary doesn't get in the way.
Note that no matter what $x$ we choose, for a simplex $\sigma: [0,1] \to S^1$ with $\sigma(0) = \sigma(1) \neq x$, we have $\deg_x(\sigma) = \text{Wind}(\sigma)$. 
