Let $\Gamma$ be a normal subgroup of finite index of the modular group $PSL(2,\mathbb{Z})$.
Let $\mathbb{H}$ be the upper half-plane.
Let $\phi$ be a cusp form of weight $2$ for $\Gamma$. Then $\omega=Re(\phi(z)dz)$ (the real part of $\phi(z)dz$, if $\phi(z)dz=(u(x,y)+iv(x,y))(dx+idy)$, then $\omega=u(x,y)dx-v(x,y)dy$) is a harmonic $1-$ form on $\mathbb{H}$. It induces a harmonic $1-$ form on $\Gamma\backslash\mathbb{H}$ also denoted by $\omega$.
Let $dv$ be the normalised volume element on $\Gamma\backslash\mathbb{H}$.
If case we have the following integral $$\int_{\Gamma\backslash\mathbb{H}}\|\omega\|^2dv$$ Which norm do they mean ($\|\omega\|$)?
In your paper they are looking at $$\omega =f(z)dz= f(z)dx+if(z)dy$$ for $f$ a weight 2 cusp form. Then $$\omega\wedge \overline{\omega}= (f(z)dx+if(z)dy)\wedge (\overline{f(z)}dx-i\overline{f(z)}dy) $$ Using bilinearity and $dx\wedge dy=-dy\wedge dx,dx\wedge dx=0$ you get that is $$ = -2i|f(z)|^2\ \ dx\wedge dy= -2i y^2|f(z)|^2\ \ \frac{dx\wedge dy}{y^2} $$ $\frac{dx\wedge dy}{y^2}$ is the area $2$-form and you can check that $y^2|f(z)|^2$ is $\Gamma$ invariant so that $$\|f(z)dz\|^2=\int_{\Gamma\backslash \Bbb{H}} y^2|f(z)|^2\ \ \frac{dx\wedge dy}{y^2}$$ makes sense.
This gives a norm on $S_2(SL_2(\Bbb{Z}))$. For harmonic 1-forms the natural norm is $$\sup_{f\in S_2(SL_2(\Bbb{Z})),\|f\|^2\le 1} |\int_{\Gamma\backslash \Bbb{H}} \omega \wedge \Re( f(z) dz)|$$ Which relates to the Petersson inner product.