This is Definition 1.3.16, Definition 1.3.17 from Scharlau's book Quadratic and Hermitian Forms
My question is what is well defined upto squares mean in definition 3.17 Also in second line of 3.17 det(V,b) is element of $K^.$/$K^.2$ what is meaning of that? Say we have real number field with exclusion of 0 as $K^.$ for the bilinear form b then; what it means is if it is nonsingular then det(V,b) is all elements excluding squares. Is it right or wrong?
The notation $K^\cdot/(K^\cdot)^2$ simply means quotient group of $K^\cdot$ by the subgroup $(K^\cdot)^2$ consisting of all squares. In other words, these are the equivalence classes of $K^\cdot$ if you take equivalence relation $$k\sim k' \Leftrightarrow k'k^{-1}\in(K^\cdot)^2.$$
You can find more information in the Wikipedia article Square class. Some examples are listed on GroupProps.
In the fields you encounter most frequently this is quite simple. Every complex number has a square root, so if $K=\mathbb C$ you only have one class. If $K=\mathbb R$ you get two classes, positive and negative numbers (squares and non-squares). Similarly in finite fields you have at most two classes, see here: The number of elements which are squares in a finite field.
The phrase "well defined up to square" means exactly what is written in the following sentence: "$\det B$ and $\det B'$ differ only by element of $(K^\cdot)^2$". The other way to put is is that you can assign in this way to $(V,b)$ the equivalence class od $\det B$ w.r.t. to the equivalence relation described above.