It's maybe also important to note that we deal with two different kinds of vectors here when used for geometry (and the kinematics of physics is directly just applied geometry; the rest of physics is also geometry, but on a somewhat more abstract level :-)).
Usually you start to use vectors in (Euclidean) geometry in the sense of "polar vectors". They always have to do with translations or a direction along something is moving. The most direct example is a vector defined by two points in Euclidean space ##\overrightarrow{AB}##, which you can use to depict the translation of the point ##A## to the point ##B## along the straight line connecting the two points. That's where the name "vector" comes from (in Latin "vehere" means "to drag" or "to move").
Now you describe the motion by defining some "origin" ##O## and each point ##P## is then uniquely defined by the position vector ##\vec{x}=\overrightarrow{OP}##, and thus you can describe the trajectory of a point particle as a function of time, and that's what we are after in classical point-particle mechanics. Time is mathematically just a parameter which parametrizes the trajectory (but with a physically well-defined measure which is given by some clock). This already implies that also time derivatives of ##\vec{x}## all are polar vectors. Particularly useful in mechanics are of course velocity ##\vec{v} = \dot{\vec{x}}## and acceleration ##\vec{a}=\dot{\vec{v}} = \ddot{\vec{x}}##, and both are polar vectors.
As was already stated in the answers above, there are also another kind of vectors, the socalled "axial vectors", where "axial" refers to the idea of a axis around which something rotates. For this by convention we use the right-hand rule: To describe a rotation of a point around an axis you point the thumb of your right hand in direction of a unit vector ##\vec{n}##. Then the (naturally) curled fingers give the sense of the rotation of the point around the axis. Now you also need the angle of the rotation ##\phi \in [0,2 \pi)##. In this sense you can define the rotation by the vector ##\vec{\phi}=\phi \vec{n}##, but here indeed the vector doesn't have anything to do with dragging something from one place to another but rather it gives a precise description in which way something rotates around the axis.
Now take as a simple example a particle moving on a circle of radius ##a## around the ##z## axis of a Cartesian right-handed basis system (i.e., the basis vectors ##\vec{e}_i## are perpendicular to each other and of unit length, i.e., ##\vec{e}_i \cdot \vec{e}_j=\delta_{ij}## and oriented relative to each other such that if you point the thumb of your right hand in direction of ##\vec{e}_1##, the index finger in direction of ##\vec{e}_2##, then the middle finger points in direction of ##\vec{e}_3##).
This motion is obviously described by
$$\vec{r}(t)=\vec{e}_1 a \cos[\phi(t)] + \vec{e}_2 \sin[\phi(t)],$$
where the angle of rotation is an arbitrary function of time. Now calculate the velocity (a polar vector!)
$$\vec{v}(t)=\dot{\vec{r}}(t)=a \dot{\phi}(t) \{-\vec{e}_1 \sin [\phi(t)] + \vec{e}_1 \cos[\phi(t)] \}.$$
On the other hand
$$\vec{e}_3 \times \vec{r}(t)= a \{-\vec{e}_1 \sin [\phi(t)] + \vec{e}_1 \cos[\phi(t)] \}.$$
So we can write
$$\vec{v}(t)=\dot{\phi}(t) \vec{e}_3 \times \vec{r}(t)=\dot{\vec{\phi}}(t) \times \vec{r}.$$
From this it's clear that the cross product describes infinitesimal rotations, because the infinitesimal change of the position due to the rotation of the point around the ##z## axis is given by
$$\mathrm{d} \vec{r} = \mathrm{d} \vec{\phi} \times \vec{r},$$
and that's why we call ##\vec{\omega} = \dot{\vec{\phi}}## the angular velocity, and it's clear that it is an axial vector as is the case for ##\vec{\phi}##.
The last question now is, how you can distinguish mathematically a polar from an axial vector. The answer is, how the vectors behave under spatial reflections. A spatial reflection at the origin of the coordinate system for a position vector is defined by ##\vec{r} \rightarrow \vec{r}'=-\vec{r}##. Since of course time has nothing to do with the reflection, by definition it doesn't change. Thus also ##\vec{v}=\dot{\vec{r}}## and ##\vec{a}=\ddot{\vec{r}}## just flip their sign under spatial reflections.
For the rotation of the particle around the ##z## axis we got
$$\vec{v}=\vec{\omega} \times \vec{r}.$$
Now ##\vec{v}## and ##\vec{r}## are polar vectors and thus both flip sign under rotations. For that to be consistent with this formula, you must define that the axial vector ##\vec{\omega}## does NOT flip its sign, i.e., under spatial reflections ##\vec{\omega} \rightarrow \vec{\omega}'=\vec{\omega}##.
So the axial vectors simply don't change at all under spatial reflections while polar vectors flip their sign.
Another way to see this is to look again at the formula for the velocity of the particle rotating around the ##z## axis. We just take the cross product with ##\vec{r}##:
$$\vec{r} \times \vec{v}=\vec{r} \times (\vec{\omega} \times \vec{r}) = r^2 \vec{\omega}-\vec{r} (\vec{r} \cdot \vec{\omega})=a^2 \vec{\omega}.$$
Now we have expressed the axial vector ##\vec{\omega}## in terms of a cross product of two polar vectors, and obviously under space reflections we have
$$\vec{r} \times \vec{v} \rightarrow \vec{r}' \times \vec{v}'=(-\vec{r}) \times (-\vec{v})=\vec{r} \times \vec{v},$$
i.e., the cross product of two polar vectors is an axial vector.
