1) Does this mean that for any particle on the rotating body the angular velocity is the same?
On a rigid rotating body, yes, the angular velocity is the same for every point in that body.
2) Does this mean that when angular momentum is described, we are technically still describing a relationship between linear velocity and mass (mv), only now the linear velocity depends on angular velocity and radial distance from axis?
In effect, yes. What you are setting up is an equation of momentum for every infinitesimal mass element of your body. You see the analogy between linear and angular momentum:
$$p = mv$$ and $$L = I\omega$$ where $I$ depends on the distribution of mass, not just on the total mass itself.
3) this would mean that linear velocity would be less for particles close to the axis of rotation, but angular velocity would be the same?
That's exactly what's happening. To visualize this, simply imagine spinning a weight fixed to a string over your head. If you spin one weight with a certain angular speed $\omega$ and then release the string, it will fly off at a certain speed. Do the same now with a shorter string but the same angular speed. The weight will fly off at a slower velocity.
4) Then why would something like a pulsar rotate faster as its matter get closer to the center?
Since angular momentum is conserved, decreasing the moment of inertia increases angular speed: $L = I \omega$. As an analogy, consider a pirouette of an ice skater. If the ice skater has her arms outstretched (big moment of inertia) and rotates at a certain angular speed $\omega$, after she pulls in her arms, she will spin at a faster rate. This is because the angular momentum she had before is the same as afterwards.
All in all, angular speed/momentum follows a neat analogy with linear speed/momentum.
I'll try to give here a very partial answer to the question. Not sure whether it is interesting in itself, but it might provide a hint for further development. Possibly its place would be within a comment, but comments are limited in length and it wouldn't fit.
Let's define $\alpha (t_0,t)$ as the $\alpha (t)$ from the question related to the given $t_0$. We identify the rotation vector with the rotation itself. For $t_i = t_0 + i\ dt$ and $t = t_0 + n\ dt$, we have by simple composition of successive rotations
$$\alpha (t_0,t) = \prod_{i=n-1}^{0} \alpha (t_i,t_{i+1}).$$
We know from this question/answer that $\partial_2 \alpha (t,t) = \omega(t)$ for all $t$. Using $\alpha (t,t+dt) = I+\partial_2 \alpha (t,t)\ dt + o(dt) = I+\omega(t)\ dt + o(dt)$, we have
$$\alpha (t_0,t) = \prod_{i=n-1}^{0} I+\omega(t_i)\ dt + o(dt).$$
This provides by the way a numerical method expressing $\alpha$ in terms of $\omega$.
In the very special case where all these rotations commute (example of common axis), the limit is an exponential: taking the log and the limit when $dt \to 0$, we have
$$\log (\alpha (t_0,t)) = \sum_{i=n-1}^{0} \omega(t_i)\ dt + o(dt) \to \int_{t_0}^{t} \omega(s) \ ds,$$
hence
$$\alpha (t_0,t) = \exp \left( \int_{t_0}^{t} \omega(s) \ ds \right).$$
In the general non commutative case, the log will involve Lie brackets starting with $dt^2[\omega(t_i),\ \omega(t_j)]$ (cf. Dynkin's formula), and some more courage seems required.
EDIT: according to a comment below from Keshav Srinivasan, the expression above becomes in the general non commutative case
$$\alpha (t_0,t) = \operatorname{OE}[\omega](t_0,t)= \mathcal{T} \left\{e^{\int_{t_0}^{t} \omega(s) \, ds}\right\},$$
see Ordered exponential for the definition of $\operatorname{OE}[\omega](t)$. Though this is not an exact answer to the question, as it involves rotation matrices instead of the required rotation vectors.
Best Answer
OK, I'm assuming you want the formal proof of this well known kinematics formula! So here goes:
Let the particle rotate about the axis OO' ... Within time interval $dt$ let its motion be represented by the vector $d\varphi$ whose direction is along axis obeying the right-hand-corkscrew rule, and whose magnitude is equal to the angle dφ.
Now, if the elementary displacement of particle at a be specified by radius vector $r$,
From the diagram, it is easy to see that, for infinitesimal rotation, $dr= d\varphi\times r \tag{1}$
By definition, $ω = dφ/dt$
Thus taking the elementary time interval as $dt$, all given equations surely hold!
Thus we can divide both sides of the equation $(1)$ by $dt$ which is the corresponding time interval!
So we get $dr/dt = dφ/dt \times r$ of course $r$ value won't change WRT the particle and axis, so $r$/dt is essentially $r$!
So result is, $$\boxed{v = ω \times r}$$