Because $M_4(\mathbb R) = M_2(\mathbb R) \otimes M_2(\mathbb R)$ as vector spaces (and as algebras, but we won't use this), we can replace $M_2(\mathbb R)$ by an arbitrary $4$-dimensional vector space $V$ and $M_4(\mathbb R)$ by $V \otimes V$. We can represent elements of $V\otimes V$ conveniently as $4\times 4$ matrices, where simple tensors are rank one matrices. The question is then equivalent to asking the minimum number of rank $1$ matrices it takes to write a $4 \times 4$ matrix as a sum of rank $1$ matrices. The answer is obviously $4$.

For quaternions acting by left multiplication by quaternions in the standard basis:

$$1= \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\0 & 1 \end{pmatrix} \otimes\begin{pmatrix} 1 & 0 \\0 & 1 \end{pmatrix}$$

$$i= \begin{pmatrix} 0 & -1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 \\ 0 & 0 & 1 & 0 \end{pmatrix} = \begin{pmatrix} 0 & -1 \\1 & 0 \end{pmatrix} \otimes\begin{pmatrix} 1 & 0 \\0 & 1 \end{pmatrix}$$

$$j= \begin{pmatrix} 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \otimes\begin{pmatrix} 0 & -1 \\1 & 0 \end{pmatrix}$$

$$k= \begin{pmatrix} 0 & 0 & 0 & -1 \\ 0 & 0 & -1 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \otimes\begin{pmatrix} 0 & -1 \\1 & 0 \end{pmatrix}$$

Because only two terms appear on the right side, the matrices clearly have rank at most $2$ at tensor product, and some obtain rank exactly $2$, so the answer is two.

For the last question, the answer is isomorphic to the analogous answer for writing an element of $V \otimes V$ as $v_1 \otimes v_2 - v_2 \otimes v_1$ for $V$ a four-dimensional vector space. The condition is given by a skew-symmetry condition (i.e. 10 linear conditions) plus a Pfaffian condition (a quadratic condition). More precisely the general such matrix can be written as

$$ \begin{pmatrix} 0 & a & -a & 0 \\ b & c & d & e \\ -b & -d & -c & -e \\ 0 & f & -f & 0 \\ \end{pmatrix}$$

such that $af-be+cd=0$