In Wikipedia both statement is present, which one is true?
In one hand, here stands:
def 1.
Thus the (real) spinors in three-dimensions are quaternions, and the action of an even-graded element on a spinor is given by ordinary quaternionic multiplication.
while on the other hand, here stands:
def 2.
These are 2×2 matrices with complex entries, and the two-component complex column vectors on which these matrices act by matrix multiplication are the spinors.
It's clear that both quaternions and elements of $\mathbb C^2$ can be represented by four real numbers, but I don't see that these two things above are the same.
According to the context of def 1., "even-graded elements" and "quaternions" are the same (the elements of the space spanned by the scalars and the i-multiples of the Pauli-matrices). Then, quaternions are the same as spinors? Then "the action of an even-graded element on a spinor" can also be formulated as "the action of a spinor on a spinor"?
If the above two definitions are the same, then what is the correspondence between the elements of the column vectors in def 2. and the elements of the i-multiples of the Pauli matrices so that the two kinds of multiplication result in the same?
Spinors are elements of a certain representation, so "action of a spinor on a spinor" does not make much sense (even if it's obtained by a formally correct substitution). The right way to think about this is that multiplication of quaternions gives an action of a "quaternion seen as an even Clifford element" on a "quaternion seen as a spinor". There is an inherent distinction between what is acting and what is acted on.
The slight problem here is that two description talk about different actions. The description with the quaternions talks about a representation of the even Clifford algebra $\mathrm{Cl}^0_{3,0}$, which happens to be isomorphic to $\mathbb{H}$, and the description with $\mathbb{C}^2$ and Pauli matrices talks about representation of the Lie algebra $\mathfrak{so}_3$, which happens to be isomorphic to $\mathfrak{su}_2$.
Let's see how to disentangle this. In the end, we want spinors to be related to the rotations of the three-dimensional space, or, more specifically, the infinitesimal rotations, elements of the Lie algebra $\mathfrak{so}_3$ (spinors are strange precisely because of the fact that integrating this action over the full revolution we don't get to where we started, but change the sign).
Rotations are represented by orthogonal matrices $A$, that is, $AA^T = 1$. Taking the tangent space we obtain the Lie algebra $\mathfrak{so}_3$, which consists of antisymmetric matrices (intuitive explanation: for $1 + \varepsilon a$ to be an "infinitesimal rotation", we need $(1 + \varepsilon a)(1 + \varepsilon a)^T=1+\varepsilon(a+a^T)+\varepsilon^2aa^T \approx 1 + \varepsilon(a + a^T) = 1$, so $a = -a^T$). The basis of this space consists of thee matrices $$X = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & -1 & 0\end{bmatrix}, Y = \begin{bmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ -1 & 0 & 0\end{bmatrix}, Z = \begin{bmatrix} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}$$ They have the commutation relations $[X, Y] = Z, [Y, Z] = X, [Z, X] = Y$. There are other objects that have the same commutation relation. In particular, the multiples of Pauli matrices $X = \frac{-i \sigma_x}2$, $Y = \frac{-i\sigma_y}2$, $Z = \frac{-i \sigma_z}2$, and the multiples of quaternionic units $X = \frac{i}{2}, Y = \frac{j}{2}, Z = \frac{k}{2}$.
The correspondence you are asking for consists of two parts:
On the Lie algebra side, a pure quaternion $v = bi + cj + dk$ (seen as an infinitesimal rotation) corresponds to a matrix $r_v = b(-i\sigma_x) + c(-i\sigma_y) + d(-i\sigma_z) \in \mathbb{C}^{2\times 2}$ (seen as an infinitesimal rotation). This is because $(i,j,k)$ and $(-i\sigma_x, -i\sigma_y, -i\sigma_z)$ have the same commutation relations.
On the spinor side, a quaternion $q = a + bi + cj + dk$ (seen as a spinor) corresponds to the vector $s_q = \begin{bmatrix} a-di \\ c-bi \end{bmatrix} \in \mathbb{C}^2$ (seen as a spinor). We can find this by associating $1 \leftrightarrow \begin{bmatrix}1 \\ 0\end{bmatrix}$ and seeing where it will go when multiplied by $-i\sigma_{*}$.
In the end, we have $r_v s_q = s_{vq}$.