$\renewcommand{\ket}[1]{\left \lvert #1 \right \rangle}$
$\renewcommand{\bra}[1]{\left \langle #1 \right \rvert}$
We can see how decoherence really works, why it messes up superposition states, and why it's particularly prone to messing up states of large objects all through a very simple example $^{[a]}$.
Single two-level system
Suppose we have a quantum system $S$ with two possible states.
$S$ could be a cat and the states could be $\left \lvert \text{alive} \right \rangle$ and $\left \lvert \text{dead} \right \rangle$, but for the sake of generality we label the states as
$$\ket{\uparrow} \quad \text{and} \quad \ket{\downarrow} \, .$$
Coherent case
Now suppose $S$ is in state $\ket{y}$ defined as
$$\ket{y} \equiv \left( \ket{\uparrow} + i \ket{\downarrow} \right) / \sqrt{2} \, .$$
This is a perfectly happy superposition state.
It's density matrix is
$$ \rho_S = \ket{y} \bra{y} = \frac{1}{2} \left(
\ket{\uparrow} \bra{\uparrow}
+ \ket{\downarrow} \bra{\downarrow}
- i \ket{\uparrow} \bra{\downarrow}
+ i \ket{\downarrow} \bra{\uparrow}
\right)
= \frac{1}{2} \left[ \begin{array}{cc} 1 & -i \\ i & 1 \end{array} \right]
= \frac{1}{2} \left( \text{Id} + \sigma_y \right) \, ,
$$
where in the matrix representation we've ordered the states $\{ \ket{\uparrow}, \ket{\downarrow} \}$.
We can think of this state as a spin pointed along the $y$ axis (hence the symbol $\ket{y}$).
The first two terms are the classical terms (diagonal in the matrix representation) and the other two are the so-called "coherences" (off-diagonal) which disappear via decoherence processes, as we show below.
If we were to prepare $S$ in state $\ket{y}$ many times and each time measure it along the $z$ axis we would get a random sequence of results where half of them are $\uparrow$ and half are $\downarrow$.
Naively you might think that this means that our preparation procedure is giving us a normal "classical" probability distribution where half of the time we prepared $\ket{\uparrow}$ and half of the time we prepared $\ket{\downarrow}$.
However, we can see that this is not true if we rotate $S$ about the $x$ axis and then measure it along the $z$ axis.
The operator for the rotation is
$$
U = \cos(\theta / 2) \, 1 + i \sin(\theta / 2) \, \sigma_x
= \left[ \begin{array}{cc} \cos(\theta/2) & i \sin(\theta/2) \\ i \sin(\theta / 2) & \cos(\theta / 2) \end{array} \right]
$$
and the density matrix after the rotation is
$$
U \rho_S U^\dagger =
\frac{1}{2} \left[ \begin{array}{cc} 1 - \sin(\theta) & -i \cos(\theta) \\ i \cos(\theta) & 1 + \sin(\theta) \end{array} \right]
= \frac{1}{2} \left( \text{Id} - \sin(\theta) \sigma_z + \cos(\theta) \sigma_y \right) \, .
$$
As you can see, for a given angle $\theta$, the probability to find the system in $\ket{\uparrow}$ is $(1/2)(1 - \sin(\theta))$, i.e. it depends on how much we rotated.
Another way to say this is that
$$
\langle \sigma_z \rangle_{U \rho_S U^\dagger} = - \sin(\theta) \, ,
$$
i.e. the expectation value of $\sigma_z$ oscillates as we rotate the system.
This makes perfect sense if you think of the two level system as an arrow oriented in 3D space (e.g. a spin): as we rotate the system about the $x$ axis its projection about the $z$ axis oscillates.
So far, nothing about this example tells us anything about decoherence or why it's hard to make big Schrodinger cat states, so now let's get to that.
Incoherent case
Suppose $S$ interacts with some other two level system $E$.
The letter $E$ stands for "environment" which will make sense later.
Suppose the state of the combined $S + E$ system is $^{[b]}$
$$
\left( \ket{\uparrow}\ket{\downarrow} + \ket{\downarrow}\ket{\uparrow} \right) / \sqrt{2}
$$
where the first ket labels the state of $S$ and the second ket labels the state of $E$.
Now the critical part: what happens if we now do the rotate-and-measure experiment described above on system $S$ without doing anything, including measurement, to $E$?
Experimentally, when we try this in the lab, we find that there is no oscillation in the probability to find $S$ in $\ket{\uparrow}$ as a function of $\theta$!
This is decoherence.
To describe this mathematically we look at the density matrix of the $S+E$ system.
The state of the total system is
$$
\rho_{S+E} =
\left[
\begin{array}{cccc}
0&0&0&0 \\
0 & 1/2 & 1/2 & 0 \\
0 & 1/2 & 1/2 & 0 \\
0&0&0&0
\end{array}
\right]
$$
where we've ordered the states $\{ \ket{\uparrow}\ket{\uparrow}, \ket{\uparrow}\ket{\downarrow}, \ket{\downarrow}\ket{\uparrow},\ket{\downarrow}\ket{\downarrow} \}$.
To predict the behaviour of experiments done on $S$ alone, we take the trace of $\rho_{S+E}$ over the part of the space belonging to $E$ $^{[c]}$.
Doing this gives
$$
\tilde{\rho}_S \equiv \text{Tr}_E \left( \rho_{S+E}\right) =
\frac{1}{2} \left[
\begin{array}{cc}
1 & 0 \\ 0 & 1
\end{array}
\right]
= \frac{1}{2} \left( \ket{\uparrow}\bra{\uparrow} + \ket{\downarrow}\bra{\downarrow} \right) = \frac{1}{2}\text{Id} \, .
$$
The off-diagonal terms are gone - we have a purely classical state!
If we now rotate $\tilde{\rho}_S$ by any rotation operator $U$ we find that
$$
U \tilde{\rho}_S U^\dagger = \frac{1}{2} \left[ \begin{array}{cc} 1&0\\0&1 \end{array} \right] = \tilde{\rho}_S $$
and
$$ \langle \sigma_z \rangle_{U \tilde{\rho}_S U^\dagger}
= \text{Tr}_S (U \tilde{\rho}_S U \sigma_z) = 0 \, .$$
Rotations no longer do anything - there's no oscillation and the expectation value of $\sigma_z$ is always zero regardless of rotation angle.
This is really, really interesting.
Previously we said that a single isolated two level system can be thought of like a spin particle: it's always pointing in some direction in space, so even if measurements along the $z$ axis give half up and half down, if you rotate the spin and measure, you see oscillation.
On the other hand, we just showed that if we let the two level system interact with something else ($E$), the combined system can be left in a state such that the original two level system ($S$) doesn't exhibit that oscillation.
What we have just seen is the essence of quantum decohrence.
If a quantum system $S$ becomes entangled with its surrounding environment $E$, then $S$ can lose its quantum nature.
Of course, if we don't ignore the environment $E$ and instead include it in our measurements, then we would observe the full quantum properties of the combined system.
In other words, decoherence is just lack of knowledge of the complete system.
If $E$ is really big then keeping track of all its degrees of freedom and measuring them in a controlled way is just impossible.
That is the essence of why making big Schrodinger cats is hard; if the system $S$ is big it interacts with more environmental degrees of freedom and so observing quantum effects is very difficult.
For something a large as a speck of dust interacting with air molecules, the time it takes for the decoherence to kill any off diagonal elements in the density matrix is incredibly small $^{[d]}$.
Interestingly though, some fairly large systems can be sufficiently isolated from their environments such that they exhibit quantum properties long enough times to be useful; this is, for example, a large fraction of what goes into building a quantum computer.
Large system
So far we showed what decoherence is, and in particular how it makes a quantum system appear classical.
To recap, decoherence happens when your system $S$ interacts with the environment $E$; if you don't have access to the environment degrees of freedom, then $S$ can lose its quantum interference properties and appear classical.
In the example we gave, we saw that a single two level system interacting with another one can appear classical.
We will now, via an illustrative extention of the same example, that a large system is more prone to decoherence.
Suppose $S$ consists of three two level systems in an initial state $\ket{\uparrow \uparrow \uparrow}$ with density matrix
$$\rho = \ket{\uparrow \uparrow \uparrow} \bra{\uparrow \uparrow \uparrow} \, .$$
Note that there's a sort of redundancy here: we have three separate spins which can be thought of as collectively representing a single spin up.$^{[e]}$
Like in the single particle case, we can measure the projection of the spin along the $z$ axis, but in this case we use the three-particle operator
$$Z^{(3)} \equiv \left( \sigma_z \otimes \sigma_z \otimes \sigma_z \right) \, .$$
Coherent case
Like before, if we rotate all three spins and measure the average of $Z^{(3)}$ we get a sinusoidal dependence on the rotation angle.
In particular, if we rotate each spin by an angle $\theta$ about the $x$ axis, then we get
$$\langle Z^{(3)} \rangle_{U \rho U^\dagger} = \cos(\theta)^3 \, .$$
Incoherent case
Now consider what happens if just one of our spins interacts with the environment.
Suppose the middle spin interacts with the environment such that the initial state $\ket{\uparrow \uparrow \uparrow} \ket{\downarrow}$ (here the separate second ket with one arrow represents the environment) becomes
$$
\left(
\ket{\uparrow \uparrow \uparrow}\ket{\downarrow}
+ \ket{\uparrow \downarrow \uparrow}\ket{\uparrow}
\right) / \sqrt{2} \, .
$$
Writing out the complete four particle density matrix would be tedious and unenlightening.
However, the reduced density matrix of the first three particles is
$$\tilde{\rho}_S =
\frac{1}{2} \left(
\ket{\uparrow \uparrow \uparrow}\bra{\uparrow \uparrow \uparrow}
+ \ket{\uparrow \downarrow \uparrow}\bra{\uparrow \downarrow \uparrow}
\right) \, .
$$
Note that we have a diagonal density matrix just like we did in the single particle incoherent case.
With this density matrix, the expectation value of $Z^{(3)}$ following a rotation of all spins by $\theta$ is
$$\langle Z^{(3)} \rangle_{U \tilde{\rho}_S U^\dagger} =
\frac{1}{2} (
\underbrace{\cos(\theta)^3}_{\text{from } \ket{\uparrow \uparrow \uparrow}} + \underbrace{-\cos(\theta)^3}_{\text{from } \ket{\uparrow \downarrow \uparrow}}
) = 0 \, .
$$
Here again we've lost the oscillation, and it only took a single spin interacting with the environment to do it.
That is why making large Schrodinger cats is hard.
Notes
$[a]$: This is a simplified version of an example from the introductory chapter of my PhD thesis (pdf).
$[b]$: This state can be realized if we start with the system in $\ket{\uparrow}\ket{\downarrow}$ and subject the system to the Hamiltonian $H=\sigma_+ \sigma_- + \sigma_- \sigma_+$ for the proper amount of time such that the propagator is
$$U = \left [ \begin{array}{cccc}
1&0&0&0\\
0&\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}&0\\
0&\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}&0\\
0&0&0&1 \end{array}
\right ] \, .
$$
$[c]$: Note that, like any other theoretical description, this procedure is justified because it reproduces the results of experiments.
$[d]$: I don't remember the numbers but see Schlosshauer's book for a calculation.
$[e]$: This kind of redundancy is critical in classical machines.
For example, a memory bit in a classical computer could be represented by current in a large number of conduction channels in a transistor; if any one of those channels were to change state, that's such a tiny fraction of the total current that the logical state of the transistor is preserved.
This redundancy gives classical computers their robustness against errors on the microscopic level.
