Theorem states that every problem in $\mathsf{NP}$ can be reduced to another $\mathsf{NP}$-complete problem in polynomial time. Also you can only make a problem polynomially longer using polynomial time.
Thus SAT of length $n$ can be reduced to knapsack with length $n^k$ for some fixed $k\in\mathbb{R}^+$. We don't need to show that reduction, we only need to know that it exists.
Knapsack has quasipolynomial algorithm, in other words running time is $O(\alpha^{\log^p q})$ where $\{p,\alpha\}\in\mathbb{R}^+$ are some fixed numbers and $q$ is length of knapsack problem.
Now assuming we have converted SAT to knapsack we get runtime $O(\alpha^{\log^p n^k})=O(\alpha^{k^p\log^p n})=O(\beta^{\log^p n})$ where $\beta\in\mathbb{R}^+$ is some fixed number. So, the runtime is quasipolynomial.
Why doesn't this simply negate ETH?