On reducing the hardness of CNF-SAT to k-Clique CNF-SAT refers to the following problem:
  
  Given a boolean formula $\phi$ in conjunctive normal form, does there
  exist an assignment to the variables that satisfies $\phi$.


There are several parameters that one can associate with $\phi$:


$n$ will denote the number of variables in $\phi$.
$m$ will denote the number of clauses in $\phi$.
$N$ will denote the number of variable occurrences in $\phi$.



  Claim: For every fixed $k$, we can reduce an arbitrary instance $\phi$ of CNF-SAT to an instance $G$ of $k$-Clique with roughly
  $2^{\frac{N}{k}}$ vertices and $2^{\frac{2 N}{k}}$ edges.
  
  Proof Idea: Any clause in $\phi$ can be split into two clauses by adding an
  additional variable.  The positive literal gets added to one of the
  new clauses and the negative literal gets added to the other new clause.
  
  We make at most $k$ splits to the clauses of $\phi$ to get a new
  formula $\phi'$ such that:
  
  
  $\phi'$ has at most $N+2k$ variable occurrences.
  $\phi'$ can be expressed as $\bigwedge_{i\in[k]} \phi_{i}'$ where each $\phi_{i}'$ is a CNF formula with at most $\frac{N}{k} + 2$
  variable occurrences.
  $\phi$ is satisfiable if and only if $\phi'$ is satisfiable.
  
  
  We can look at each $\phi_{i}'$ independently and build up a list of
  all the satisfying assignments.  There are at most $2^{\frac{N}{k}+2}$
  satisfying assignments for each $\phi_{i}'$.
  
  The satisfying assignments for the $\phi_i'$'s will represent the
  vertices of the graph $G$.  In total, $G$ will have at most $k \cdot 2^{\frac{N}{k}+2}$ vertices.
  
  Vertices $v$ and $u$ are adjacent in $G$ if the following are true:
  
  
  $v$ and $u$ represent assignments for $\phi_i'$ and $\phi_j'$ respectively where $i \neq j$.
  For the variables that are in both $\phi_i'$ and $\phi_j'$, the assignments for $v$ and $u$ agree on their respective bit values.
  
  
  Now, if a $k$-Clique exists in $G$, then we get an assignment for each
  $\phi_i'$ where no two assignments disagree on bit values for the same
  variables.  As a result, these assignments can be joined to form a
  satisfying assignment for $\phi'$.
  
  Conversely, if a satisfying assignment exists for $\phi'$, then we can
  remove variables from this assignment to get a satsfying assignment
  for each of the $\phi_i'$'s where no two assignments have any bit
  mismatches making them form a $k$-Clique in $G$.  $\square$


Using fast matrix multiplication, we can solve $k$-Clique in $O(n^{.792 k})$ time.  Therefore, we can solve CNF-SAT in $2^{.792 N}$ time.  In fact, there is a trivial algorithm for solving CNF-SAT in $2^{.5 N}$ time.

However, it is not known if we can solve CNF-SAT in $poly(N) \cdot 2^{(1-\epsilon) \cdot n}$ time.

Question: Does there exist a constant $c$ such that for every fixed $k$, we can reduce an arbitrary instance $\phi$ of CNF-SAT to an instance $G$ of $k$-Clique with roughly $2^{\frac{c \cdot n}{k}}$ vertices and $2^{\frac{2 \cdot c \cdot n}{k}}$ edges.

Relevant Links:


Exponential Time Hypothesis: https://en.wikipedia.org/wiki/Exponential_time_hypothesis
Algorithms for $k$-Clique: http://theory.stanford.edu/~virgi/combclique-ipl-g.pdf
Sparsification Lemma: https://cseweb.ucsd.edu/~russell/ipz.pdf
I don't know the answer to your specific question (it seems related to the question of whether W1=W[2]). 

But the algorithm you give in your question is subsumed by several other results. Using your definition of $N$, CNF-SAT is basically solvable in $O(1.1279^N)$ time, as in the paper by Wahlstrom (link goes to a google scholar page of papers that cite it). In particular, Wahlstrom shows that if $d$ is the average number of occurrences of a variable in the CNF, and no variable appears at most once, then CNF-SAT can be solved in $O(1.1279^{n(d - 2)})$ time. (There are other references with similarly strong running times.)

Update: Another reference is: 

Hirsch, E.A.: New worst-case upper bounds for SAT. J. Autom.
Reason. 24(4), 397–420 (2000)

This gives faster algorithms in terms of $m$ and your $N$ (which is typically called $\ell$). Namely, Hirsch gets $2^{0.30897m}$ and $2^{0.10299\ell}$ time.