Computational Model in SETH
SETH says that for all $\delta &lt; 1$ there is a $k$ such that $k$-SAT requires $2^{\delta n}$ time to be solved in the worst-case. The computational model is generally taken to be the random-access machine or pointer machine model, which allows for $O(\log N)$ time access to a storage of $N$ items, and is generally assumed to also be probabilistic with bounded error. 

As far as I know, it is open whether $2^{\delta n}$ time algorithms on such a model can be translated into two-tape Turing machines running in $2^{\delta n} \cdot poly(n)$ time. Nevertheless, proving that such a translation is not possible would separate multitape Turing machines from random access machines, which would have several very intriguing implications. For one, it would prove that SAT is not solvable in quasi-linear time on multitape Turing machines (because, if SAT could be solved with such multitape machines, then random-access machines can be efficiently simulated with multitape Turing machines). Note that many computational primitives (such as sorting, circuit evaluation, simple dynamic programming) can be implemented efficiently on multitape Turing machines. One relevant reference for these issues is Regan, "On the Difference Between Turing Machine Time and Random-Access Machine Time." 

To address your specific questions: no, a multitape Turing machine is not automatically implied here, but yes, such an "algorithm" for SAT (under the usual random-access model) would break SETH.