Finite spinor transformation

Based on the definition of $I^{munu}$ you gave in the comment, the result simply comes by expanding the sum over repeated indices. Since the only elements of $I$ that are different from 0 are $I^{01}$ and $I^{10}$, then $sigma_{munu}I^{munu} = sigma_{01} - sigma_{10}$. Now because $sigma$ is antisymmetric $sigma_{01} - sigma_{10}=2sigma_{01}$. Maybe it's easier if you don't really think of tensors as matrices, but just as collections of elements labelled by indices. In this case $sigma$ is a collection of matrices, and $I$ is a collection of numbers, so this particular tensor product is a linear combination of matrices.

As for the second question, assume our convention for the Minkowski metric is $eta_{munu}=diag(1,-1,-1,-1)$. In this case, when going from $A^{munu}$ to ${A^mu}_nu$ elements whose second index is "space-like" index (i.e. 1, 2 or 3) will get multiplied by -1. You can convince yourself of this by just writing down ${A^mu}_nu=A^{murho}eta_{rhonu}$, fixing the indices $mu$ and $nu$ and expanding the sum over repeated indices for several couples of indices. So if you want to visualize these tensors as matrices, $$A^{munu}=begin{pmatrix} A^{00} & A^{01} & A^{02} & A^{03} A^{10} & A^{11} & A^{12} & A^{13} A^{20} & A^{21} & A^{22} & A^{23} A^{30} & A^{31} & A^{32} & A^{33} end{pmatrix},qquad {A^mu}_nu=begin{pmatrix} A^{00} & -A^{01} & -A^{02} & -A^{03} A^{10} & -A^{11} & -A^{12} & -A^{13} A^{20} & -A^{21} & -A^{22} & -A^{23} A^{30} & -A^{31} & -A^{32} & -A^{33} end{pmatrix} $$