Let us start with the so-called Gagliardo-Nirenberg Inequality in $n$ dimensions, $$ \Vert u\Vert_{L^{n/(n-1)}(\mathbb R^n)}\le c_n\Vert \nabla u\Vert_{L^{1}(\mathbb R^n)}, $$$$ \Vert u\Vert_{L^{n/(n-1)}(\mathbb R^n)}\le c_n\Vert \nabla u\Vert_{L^{1}(\mathbb R^n)}, \tag{GN}$$ an inequality that can be applied to your function so that you get in two dimensions $$ \Vert u\Vert_{L^{2}(\mathbb R^n)}\le c_2\Vert \nabla u\Vert_{L^{1}(\mathbb R^n)}, $$$$ \Vert u\Vert_{L^{2}(\mathbb R^n)}\le c_2\Vert \nabla u\Vert_{L^{1}(\mathbb R^n)}, \tag{$\ast$}$$ and applying this to $u=v^2$, you obtain $$ \Vert v\Vert_{L^{4}(\mathbb R^n)}^2\le 2c_2\Vert v\nabla v\Vert_{L^{1}(\mathbb R^n)}\lesssim\Vert v\Vert_{L^{2}(\mathbb R^n)}\Vert \nabla v\Vert_{L^{2}(\mathbb R^n)}. $$ This is true for $v\in C^1_c(\Omega)$ and consequently by density in $H^1_0(\Omega)$ (which does not contain any constant non-zero function).
On periodic functions in $\mathbb R^2$: it is enough to prove $(\ast)$, but some condition must be obviously imposed. Writing for instance $$ u(x,y)=\sum_{k,l}e^{2π i(kx+ly)}\hat u(k,l), $$ we assume that $ \forall k,\ \sum_{l}\hat u(k,l)=0 ,\quad \forall l,\ \sum_{k}\hat u(k,l)=0. $ Then we can write $$ u(x,y)=\int_0^x\partial_1 u(s,y) ds=\int_0^y\partial_2 u(x,t) dt, $$ and we get $(\ast)$ by integrating wrt $x,y$ the inequality $$ \vert u(x,y)\vert^2\le \iint_{[0,1]^2}\vert\partial_1 u(s,y)\vert\vert \partial_2 u(x,t)\vert dsdt. $$ N.B. The proof of the Gagliardo-Nirenberg inequality (GN) in three or more dimensions is much more difficult, but mutatis mutandis, assuming as above the vanishing of some partial sums of the Fourier coefficients, we can get (GN) for periodic functions.