That certainly depends on what exactly you mean. I take your question as "how do you see that the (non-relativistic) electron spin (or more generally, Spin-1/2) is described by the Pauli matrices?"
Well, to start, we know that measuring the electron spin can only result in one of two values. From this we see that we need matrices of at least dimension 2. The simplest choice is then of course exactly dimension 2.
Moreover the spin is an angular momentum, and thus described by three operators obeying the angular momentum algebra: $[L_i, L_j] = \mathrm{i} \hbar\epsilon_{ijk} L_k$. This together with matrix dimension 2 basically restricts the choice to sets of three matrices which are equivalent to $\hbar/2$ times the Pauli matrices (the freedom of choice of those matrices corresponds to the freedom to use three arbitrary orthogonal directions as $x$, $y$ and $z$ direction).
So now why choose from those equivalent choices exactly the Pauli matrices? Well, there's always one measurement direction which is represented by a diagonal matrix; this makes calculations much easier. Of course it makes sense to choose the matrices in a way that this direction is one of the coordinate directions. By convention, the $z$ direction is chosen. This ultimately fixes the matrices.
Since each $\sigma_i$ is a scalar multiple of a Lie bracket of other finite matrices, each $\sigma_i$ must be traceless. So straight away we know:
$$\sigma_i=\left(\begin{array}{cc}a&b\\c&-a\end{array}\right)\tag{1}$$
and $\sigma_i^2=\mathrm{id}$ then yields $a^2 + b\,c=1$.
The eigenvalues of any matrix of the form in (1) with $a^2 + b\,c=1$ are $\pm\sqrt{a^2+b\,c} = \pm1$. Therefore, for any set of matrices we find fulfilling all the given relationships, we can do a similarity transformation on the whole set and thus (1) diagonalize any member of the set we choose whilst (2) keeping all the required relationships intact. Exercise: Prove that the given relationships (Lie brackets and $\sigma_i^2=\mathrm{id}$) are indeed invariant under any similarity transformation.
Thus, without loss of generalness, we can always choose one of the set to be:
$$\sigma_z=\left(\begin{array}{cc}1&0\\0&-1\end{array}\right)\tag{2}$$
So now work out the Lie bracket of $\sigma_z$ and $\sigma_x = \left(\begin{array}{cc}a_x&b_x\\c_x&-a_x\end{array}\right)$: result must be $2\,i\,\sigma_y$ and so we get:
$$\sigma_y = \left(\begin{array}{cc}0 & -i\,b_x \\i\,c_x & 0 \\\end{array}\right)\tag{3}$$
But given $\sigma_y^2=1$ we get $b_x\,c_x=1$ whence $a_x=0$ (since $a_x^2 + b_x\,c_x=1$). So our remaining two matrices are of the forms:
$$\sigma_x = \left(\begin{array}{cc}0 & b_x \\\frac{1}{b_x} & 0 \\\end{array}\right)$$
$$\sigma_y = \left(\begin{array}{cc}0 & -i\,b_x \\\frac{i}{b_x} & 0\\\end{array}\right)\tag{4}$$
and the remaining commutation relationships then give you the unknown constant $b_x$.
Once you have found $b_x$, we know from our comments above that any set of matrices fulfilling the required commutation relationships and $\sigma_i^2=\mathrm{id}$ is gotten from this particular set (the "standard" Pauli matrices) by a similarity transformation.
Best Answer
(I could be wrong but...) this seems like just clever rewriting with somewhat obscure notation. For instance $$ (\sigma^1+\sigma^2)^2=(\sigma^1)^2+(\sigma^2)^2+2\sigma^1\sigma^2 $$ while \begin{align} (\sigma^1)^2&=(\sigma^1_x)^2+(\sigma^1_y)^2+(\sigma^1_z)^2\, ,\tag{1}\\ \sigma^1\sigma^2&=\sigma_x^1\sigma_x^2+\sigma_y^1\sigma_y^2+\sigma_z^1\sigma_z^2\tag{2} \end{align} You can see that (2) contains bits of your $H$ so if you add and subtract smartly terms in this way you should land on your feet.
There is an implicit “vector” notation here in that $\sigma^1$ is the “vector” $\vec\sigma^1=(\sigma_x^1,\sigma_y^1,\sigma_z^1)$ so that $(\sigma^1)^2$ is $\vec\sigma^1\cdot\vec \sigma^1$ as per (1) while something like $\sigma^1\sigma^2$ is $\vec\sigma^1\cdot\vec\sigma^2$ as per (2)