I am going over this work here.
I couldn't understand the equality where the Hardy-Littlewood inequality is used. I think $\delta$ here is a weight so we can take it to be $1$ for simplicity. Would appreciate any help. I think this has something to do with Marcinkiewicz space and distribution functions but I've not been able to find a useful theorem.
I don't see any use of any Hardy-Littlewood inequality here. You have a measure space $(Q,\delta^\alpha \,dx\,dt)$, which has finite total measure. I will denote the measure by $\mu$ for simplicity. The assumption (2.7) says that $u$ is in the weak $L^{\hat q}( d\mu)$ space. Then it's just a matter of interpolation to get that $u\in L^q( d\mu)$ for every $1\le q<\hat q$. The computations use the layercake formula for $L^q( d\mu)$ norm, such as $$ \int |u|^q \,d\mu = -\int_0^\infty k^q \,d\mu(\{|u|>k\}) = q\int_0^\infty k^{q-1} \mu(\{|u|>k\})\,dk $$ See formula (9) in this post, or any real analysis book dealing with $L^p$ spaces in detail.
We have an upper estimate for $\mu(\{|u|>k\})$ to use in the above integral. But it's awkward to use it when $k$ is small, so we apply it only where $|u|>\lambda$, accounting for small values of $|u|$ separately. Then optimize for $\lambda$.