Sparsification Lemma for k-SAT and Exponential Time Hypothesis
I think your confusion might come from misquoting the statement of ETH.


  $3$-SAT instances (with $n$ variables and $m$ clauses) cannot be solved in time $poly(n)\cdot 2^{o(n)}$.


This statement is not ETH, but rather its weak (but often very useful) corollary. The ETH claims the following:


  There exists a $\delta&gt;0$ such that no algorithm with running time $poly(m)\cdot 2^{\delta n}$ can solve $3$-SAT.


Now we can actually use the Sparsification Lemma to prove the following:


  If ETH holds, then $3$-SAT instances cannot be solved in time $poly(m)\cdot  2^{o(n+m)}$.


Assume that there is an algorithm $A$ which solves $3$-SAT in time $poly(m)\cdot  2^{o(n+m)}$. For any $\delta&gt;0$, we'll construct an algorithm $B_\delta$ which solves $3$-SAT in time $poly(m)\cdot 2^{\delta n+o(n)}$. (This contradicts ETH and finishes the proof.) 

First, the algorithm $B_\delta$ takes an input formula $\phi$, applies the Sparsification Lemma with $\varepsilon=\delta$, and in time $poly(m)\cdot 2^{\delta n}$ produces $2^{\delta n}$ formulas $\phi_i$, such that $\phi$ is satisfiable if and only if $\exists i\colon \phi_i$ is satisfiable. Now the algorithm $B_\delta$ solves each of $\phi_i$ using the algorithm $A$. Note that by Sparsification Lemma the number of clauses in $\phi_i$ is $O(n)$, thus, each $\phi_i$ is solved by $A$ in time $2^{o(n)}$. Now the total running time of $B_\delta$ is $$poly(m)\cdot 2^{\delta n} + poly(m)\cdot 2^{\delta n}2^{o(n)}=poly(m)\cdot 2^{\delta n+o(n)}\; .$$