Homology - why is a cycle a boundary?

Your intuition is correct, I think. I also had the experience, when first learning this material, of wanting to understand homologies explicitly in the way that you are trying to, so I encourage you to pursue your attempt to match intuition with formal definitions.

The basic problem you observed is that often, at a technical level, one has to produce formal sums of cycles, while when thinking intuitively, one doesn't normally generate these formal sums in one's imagination. The way to reconcile this is to prove the following:

If $\alpha:[0,1] \to X$ and $\beta: [0,1] \to X$ are two 1-simplices (in any target space $X$) and $\gamma:[0,1] \to X$ is the sum of $\alpha$ and $\beta$ in the sense of addition in the fundamental group, then there is a homology between $\alpha + \beta$ (formal sum) and $\gamma$. This is easily checked, so I leave it as an exercise. (In a complete treatment of singular homology, it would appear as part of the verification of homotopy invariance, probably in some implicit manner. It is also closely related to Dylan Wilson's suggestion about verifying that homotopic cycles are homologous.) Once you've done this, you'll have more confidence that various intuitive pictures do indeed match with the more formally correct treatment.

Your construction works, but I think it's a little more complicated than it needs to be. In particular, you can just define a map from a simplex directly, without having to go through a disc or anything.

I'd do it by taking a triangle (again, let's call the vertices 0,1,2) and mapping [0,1] and [0,2] via the map that is constantly zero, and mapping [1,2] to the loop $\mu$. This can clearly be extended to the interior of the triangle ($\mathbb{R}$ is simply connected after all).

When you take the boundary, the [0,1] and [0,2] edges map to the same constant map, and thus cancel each out, so you're left with only $\mu$ in the boundary.

I agree that it's a little clunky that you need a canceling trick to turn three edges into one (and you'd need to do something similar for a 1-cycle involving two pieces), but it does end up working, and the same trick shows that any nullhomotopic loop is a boundary.

Remember that $C_2X$ does not just consist of all maps $\sigma: \Delta^2 \rightarrow X$ but also all formal sums of these. In particular, consider a map of the unit square that realizes a nullhomotopy of $\mu$. Divide the unit square into two triangles, label the vertices, orient the edges properly, and interpret the nullhomotopy as a sum of two different maps $\Delta^2 \rightarrow X$ (one might have a minus sign, actually). The boundary of this chain should be $\mu$, if all goes well.

I think, actually, this type of thing should work more generally to show that if two paths are homotopic, then they are homologous (i.e. differ by a boundary.)