The padding argument in the proof of NTIME(n) ⊆ DTIME(n^1.2) implies Σ2-TIME(n^8) ⊆ NTIME(n^9.6) In "Computational Complexity, A modern approach", Arora &amp; Barak proof the following claim (Claim 5.11.2):


  Suppose that $\mathsf{NTIME}(n) \subseteq \mathsf{DTIME}(n^{1.2})$. Then $\Sigma_2$-$\mathsf{TIME}(n^{8})  \subseteq \mathsf{NTIME}(n^{9.6})$,


which they prove in the following way:


If $L \in \Sigma_2$-$\mathsf{TIME}(n^8)$, then $\exists$ TM $M$ which runs in time $O(|x|^8)$ s.t.
$x \in L \leftrightarrow \exists u \in \{0,1\}^{c|x|^8} \forall v \in 
\{0,1\}^{d|x|^8} M(x,u,v) = 1$ 
But if $\mathsf{NTIME}(n) \subseteq \mathsf{DTIME}(n^{1.2})$, then 



  by a simple padding argument, we have a deterministic algorithm $D$ that on inputs $x, u$ with $|x| = n$ and $|u| = cn^8$ runs in time $O((n^8)^{1.2}) = O(n^{9.6})$ and returns $1$ iff there exists some $v \in \{0,1\}^{dn^8}$ such that $M(x,u,v) = 0$


I struggle to see what this "simple padding argument" might be? That is, how do we get respective deterministic algorithm $D$ if we have the TM $M$?
The language 
$$L_1 = \{(x,u) ~|~ \exists v\in \{0,1\}^{d|x|^8}M(x,u,v) = 0\}$$ belongs to $NTIME(n)$ by design, since $|v| = O(|x| + |u|) = O(|x|^{8})$. By the assumption of the theorem, $L_1$ is decidable by some deterministic $D$ in $O((|x|+|u|)^{1.2}) = O(|x|^{9.6})$ steps. The recipee for $D$ isn't important, but the existence of such algorithm $D$ leads to the last step of the proof.