It's not entirely clear (to me) what you mean by "applying an angular velocity". Angular velocity is an instantaneous property; if you only have a single angular velocity vector, you only have the first derivative of the motion at one point in time, and thus can't say anything about the orientation at other times.
I'll assume that you're assuming the angular velocity to be constant, and that by "applying" it you mean calculating the orientation at later times of a body that has the given initial orientation and undergoes a motion with the given constant angular velocity.
To do this, you need to know a) how to find the unit quaternion corresponding to a rotation vector and b) how to compose two unit quaternions in order get the unit quaternion corresponding to the composition of the corresponding rotations.
For a), let's denote the angular velocity vector by $\vec{\omega}$. Over a time $t$, a motion with this (constant) angular velocity results in a rotation that can be described by the vector $\vec{w}=\vec{\omega}t$, where the direction and magnitude of $\vec{w}$ specify the axis and angle of the rotation, respectively. This rotation corresponds to the unit quaternion
$$q=\left(\cos\frac{|\vec{w}|}{2},\frac{\vec{w}}{|\vec{w}|}\sin\frac{|\vec{w}|}{2}\right)\;.$$
For b), since applying the rotation corresponding to a unit quaternion $q$ to a three-dimensional vector $\vec{v}$ consists in conjugating a quaternion $\nu$ corresponding to $\vec{v}$ by $q$, composing rotations just corresponds to multiplying the corresponding quaternions:
$$q(\vec{v})=q\nu q^{-1}\;,$$
$$q_1(q_2(\vec{v}))=q_1(q_2\nu q_2^{-1})=q_1q_2\nu q_2^{-1} q_1^{-1}=(q_1q_2)(\vec{v})\;.$$
(This is part of what makes quaternions so useful for representing three-dimensional rotations). So now you just multiply $q$ and your initial unit quaternion, say, $q_0$. Since the initial rotation is applied first, the corresponding unit quaternion is on the inside of the nested conjugations, so it has to be the right-hand factor: $q'=qq_0$.
For more on all this, see this Wikipedia article.
Short answer
The reason that the capital $Q$'s are next to each other is because they are matrices acting in succession to transform $v$. The vector $v$ is also being transformed twice in the sequence $v\mapsto qv\mapsto qvq^\ast$, but this transformation depends on the order of multpilication (since quaternions are noncommutative), whereas when you model linear transformations by matrices on the left, they just stack up on the left. We will see that $Q$ represents $q$ and $Q^\ast$ represents $q^\ast$.
What you're seeing
You're looking at two different representations of the rotation: one as the rotation matrix $QQ^\ast$, and one as a quaternion $q$.
The first is for a column vector $v$ and two matrices you defined: the rotation is $v\mapsto Q^\ast Qv$ (You might actually want to think of as a composition of two steps: $v \mapsto Qv\mapsto Q^\ast Qv$. $Q$ will, incidentally, have to have $\det(Q)=1$ to represent a rotation.)
The second is for a vector $v$ interpreted as a quaternion with real part $0$: the rotation is $v\mapsto qvq^\ast$ where $q$ is a quaternion. (Again, you can view this as a two step process: $v\mapsto qv\mapsto qvq^\ast$. It's also important to point out that $q$ will have to be a unit length quaternion to represent a rotation, or else it does not preserve distances.)
It's important to remember that $v\mapsto qv$ and $v\mapsto vq^\ast$ are just an $\Bbb R$ linear transformations of $\Bbb H$. As such, you can fix a basis and find a matrix to represent multiplication by $q$. By choosing the right basis, the matrix produced is $Q$, and in the same basis, the matrix produced for $q^\ast$ is $Q^\ast$.
So the two sequences of mappings you see in the above two cases are really representing the same process.
(Incidentally, what about the sequence of transformations $v\mapsto vq^\ast\mapsto qvq^\ast$? Well, if you check, you'll see that $QQ^\ast=Q^\ast Q$, so doing things in the order $v\mapsto Q^\ast Qv$ still yields the same result as before :) )
Take a look at what $QQ^\ast$ looks like at Wolfram, remembering that $\det(Q)=(w^2+x^2+y^2+z^2)^2=1$. The upper left hand $3\times 3$ submatrix turns out to be a rotation matrix for $\Bbb R^3$, and the lower right hand entry is just $1$. Thus the matrix acts on the upper three rows but leaves the last row fixed. This is a clue that the first three basis vectors are where the $3$ spatial dimensions are living.
Reverse engineering the connection between matrix and quaternion
The picture we looked at in the last paragraph gives away that the authors of this representation want to represent $3$ dimensional vectors as column vectors $[x,y,z,0]^\top$. It is also likely that they want to use the "obvious" basis of $\{i,j,k,1\}$ (in that order) of quaternions to be the basis for these matrices.
Notice if $w=1$ and $x=y=z=0$, $Q$ becomes the identity matrix. Thus $w$ probably represents the real part of the quaternion, since the quaternion $1$ can represent the identity rotation.
If $w=y=z=0$ and $x=1$, we get another matrix. If you check how it acts on the coefficients of the ordered basis $\{i,j,k,1\}$, you'll find that it exactly matches left multiplication by $i\in\Bbb H$. This suggests $x$ is the coefficient for $i$ in $Q$.
Two identical analysis reveal that left multiplication by $j$ corresponds to $w=x=z=0$ and $y=1$, and that $k$ corresponds to $w=x=y=0$ and $z=1$.
Putting these things together, we have that a quaternion $w+xi+yj+zk$ with $(w^2+x^2+y^2+z^2)=1$ produces the matrix $Q$, which affects multiplication by quaternions on the left, the mapping being
$$
q=w+xi+yj+zk\mapsto\begin{pmatrix}w & -z & y & x \\ z & w & -x & y \\ -y &x &w& z\\ -x& -y & -z& w\end{pmatrix},
$$
An identical analysis reveals that $Q^\ast$ affects right multiplication by the conjugate of a quaternion on $v$. Explicitly, putting in a $1$ for $x$ and zeroes for $w,y,z$, the resulting map is right multiplication by $-i$. The mapping being given by (as you might guess)
$$
q=w+xi+yj+zk\mapsto\begin{pmatrix}w & -z & y & -x \\ z & w & -x & -y \\ -y &x &w& -z\\ x& y & z& w\end{pmatrix}
$$
Each of these matrices is producing right multiplication by the conjugate $q^\ast$.
So you can see there are two mappings at work here, both of them from the unit quaternions into $M_4(\Bbb R)$. One mapping realizes left multiplication by quaternions, the other realizes right multiplication by conjugates of quaternions.
Best Answer
You would like to sample a unit quaternion $q'$ from a normal distribution with mean $q$ and some variance $\omega$?
Background:
A rotation matrix $R_{a,b}$ can be seen as transformation which rotates a point $p$ from reference frame $b$ to the reference frame $a$: $p_a = R_{a,b}p_b$. Obviously, the same holds for unit quaterions representing rotations, so we could write $q_{a,b}$ to emphasis that this quaternion represents a rotation from frame $b$ to frame $a$ (as in $p_a = q_{a,b}\cdot p_b \cdot q_{a,b}^{-1}$).
The first question, we have to ask ourselves in which frame we'd like to sample, i.e. perturb $q_{a,b}$: frame $a$ or frame $b$? This really depends on the application. Let us simply assume $a$ for now.
Approach:
A good approach would be to sample in the tangent space at frame $a$. To be more precise, we would sample in the tangent space around the identity $\exp(\omega)$ with $\omega$ been drawn from a 3-dimensional zero-mean Normal distribution. Then, we apply this quaternion $\exp(\omega)$ to $q_{a,b}$. Thus the resulting quaterion is:
$$ q' = \exp(\omega)\cdot q_{a,b} \quad (\text{with} \quad \omega \in \mathbb{R}^3)$$
Here $\cdot$ is the quaternion multiplication, and $\exp$ the unit quaternion exponential: $\exp(\omega) = (\cos(\theta), \frac{1}{\theta}\sin(\theta)\omega )$ with $\theta = ||\omega||$.
(If we'd like to sample in frame $b$, we would do $q' = q_{a,b}\cdot \exp(\omega)$