ETH says that, for $s_k = \inf \alpha$ such that k-sat can be solved in $2^{\alpha n}$, it holds that $s_3 > 0$.
The strong exponential time hypothesis (SETH) says that $\lim\limits_{k \rightarrow \infty}s_k = 1$.
I think you mean the latter in your description.
Answering your question: The conjecture presents a lower-bound for a problem, meaning that no algorithm (in simple terms) solves all instances of $SAT$ much better than $O(2^{|V|})$[1].
However, in your question, you provided an example where a specific algorithm has a larger running time than this lower-bound for a specific class of instances.
On one hand, larger running time only makes a lower-bound stronger (it contradicts nothing).
On the other side, even if you show that for some class of instances (there are surely many of them), the problem can be solved efficiently, this does not contradict the lower-bound since it is general and your algorithm has to solve SAT more efficiently on any given instance.
On a side note, the sparsification lemma [Impagliazzo et al.], used to show that SETH is stronger than ETH reduces an instance of SAT to many instances such that in each of them we have linear number of clauses in the number of variables. This might satisfy your curiosity since it shows that SETH also has implications on instances with linear bound on the number of clauses.
Some examples of classes of formulas that can be solved efficiently are CNF formulas whose primal-, dual- or incidence-graphs have bounded values of some width parameters (tree width / branch width / etc.)
[1]: This is equivalent to the CNF-SAT conjecture. SETH is a stronger hypothesis.