What is meant by the 'linearity of quantum mechanics' in this derivation? I saw an awesome derivation of Schrodinger's equation on Wikipedia. Part of it relies on:

If the wave-function at time $t$ is given by $\Psi(t)$, then by the linearity of quantum mechanics the wave-function at time $t'$ must be given by $\Psi(t') = U(t', t)\Psi(t)$, where $U(t', t)$ is a linear operator.

What is meant by linearity of quantum mechanics here?
 A: I am supposing you are not asking for a justification for why is quantum mechanics linear, because - as some comments already pointed out - the answer would be: it is because it is, unless some future experiment shows it is not. 
This said, I am assuming you are asking why linearity in quantum mechanics implies that $\Psi(t')=U(t,t')\Psi(t)$. To keep things simple, suppose that the state at $t$ is the superposition of two states: $\Psi(t)=\psi_1(t)+\psi_2(t)$. A linear quantum evolution from time $t$ to time $t'$ means that exist a function $U_{t,t'}$ such that
$$
\Psi(t') = U_{t,t'}(\psi_1(t)+\psi_2(t))=U_{t,t'}(\psi_1(t))+U_{t,t'}(\psi_2(t)).
$$
That linear function is also supposed to satisfy $U_{t,t'}(0)=0$ and $U_{t,t'}(c\psi)=cU_{t,t'}(\psi)$, where $0$ is the zero vector and $\psi$ is an arbitrary state. The zero vector is not permissible as a quantum state, as the probability must sum up to one, and a constant multiplying an entire state can be incorporated into a normalizing factor. A function satisfying these properties is known as a linear operator as you almost certainly know. The probability must be conserved, so the norm of a quantum state must be conserved, what implies that this linear operator must be unitary. Rewriting 
$U_{t,t'}(\psi)$ as $U(t,t')\psi$ we finally arrive at
$$
\Psi(t')=U(t,t')\Psi(t).
$$
