What is the duration of the spreading sequence
When you multiply the real and imaginary parts by $\mathbf{W}_1$ and $\mathbf{W}_2$, respectively, the resulting signal is no longer $S$. It will be a matrix.  

Regarding your questions:

1- Yes, the signal after spreading has the same duration. 

2- One of the disadvantages of Walsh-Hadamard codes is that they don't have clear auto-correlation peak, which makes synchronization at the receiver side challenging. Also, they must have a dimension of $2^k$, i.e., you cannot have arbitrary dimension to support an arbitrary number of users. In your example, to make the system efficient, you need 4 users, otherwise, one user will occupy 4 times the minimum required bandwidth, which isn't spectrally efficient. 

3- Yes. This is called direct sequence (DS) CDMA.