Each row must contain two $1$'s and two $-1$'s, so identify each row with the pair of columns holding its $1$'s; each column must end up chosen by exactly two rows. There are $\binom{4}{2} = 6$ choices for row $1.$ Classify by how row $2$ overlaps row $1.$

If row $2$ uses the same pair ($1$ way), those two columns are full, so rows $3$ and $4$ must both use the complementary pair: $1$ completion. If row $2$ uses the complementary pair ($1$ way), every column has one $1$ so far, so rows $3$ and $4$ need only be a complementary pair themselves: $6$ choices for row $3,$ row $4$ forced, giving $6$ completions. If row $2$ shares exactly one column with row $1$ ($2 \cdot 2 = 4$ ways), one column is full, two have one $1,$ and one is empty; rows $3$ and $4$ must each take the empty column together with one of the two half-filled columns, so there are $2$ completions.

The total is $6\,(1 \cdot 1 + 1 \cdot 6 + 4 \cdot 2) = 6 \cdot 15$ $= 90.$