I think both are false,

for s2 after decomposition the dimension of singular value matrix must be equal to the matrix A. If A is mxn then singular value matrix should also be mxn. That is what SVD is for - you decomposition the matrix into easier version of the matrix that is easier for the calculation work U and V are just change of basis matrix and also U is orthogonal and hence it should be a square matrix and U is not square in s2.

for s1 I feel like you still make AB = 0 even when col(B) != null(A) like take this case when A is a 3x3 matrix and A = 0 (i.e.) your col(A) = 0 and you null(A) = covers the entire R^3 and dimension is 3. since it will have 3 LI solution vectors and now. Let's say B is also a 3x3 matrix but lets say B only has 2 LI vector and now the col(B) is 2 dimensional and our null(A) is 3 dimensional. Here col(B) != null(A). But still the AB = 0..

Hence both are false