Let Fn$(\kappa \times \omega,2)$ be a forcing notion. Then $2^{\aleph_0}=\kappa$ must hold in $V[G]$ but obviously
$2^{\kappa}$ cannot be $\aleph_0$.
So where is the symmetry of Fn$(\kappa \times \omega,2)$ and Fn$(\omega \times \kappa,2)$ in $V[G]$ lost?
There is no symmetry lost, these two forcings are isomorphic and in particular produce the same forcing extensions (one isomorphism arises from the obvious bijection $\omega\times\kappa\rightarrow\kappa\times\omega $ and generally $Fn (X, Y, \lambda)\cong Fn (X', Y, \lambda)$ if there is a bijection $f:X\rightarrow X'$). This forcing adds $\kappa$ many new subsets to $\omega$, but you do not know without further assumptions that $2^\omega=\kappa$ in $V [G] $ (for example this can never hold if $\kappa=\omega $) you only know for sure that $2^\omega\geq\kappa $ since there still might be a lot of "old" subsets. You observed correctly that you can also think about this forcing as adding $\omega $ many New subsets to $\kappa$ and you get $2^\kappa\geq\aleph_0$ in $V [G] $, which is trivial.