Ladner's Theorem vs. Schaefer's Theorem
You need  to understand that $\mathsf{CSP}$  problems have a structure that generic $\mathbf{SAT}$ problems do not have. I will  give  you  a  simple example. Let $\Gamma=\{\{(0,0),(1,1)\},\{(0,1),(1,0)\}\}$.   This language is  such that  you can only  express equality  and inequality between   two  variables.  Clearly any such set of constraints is solvable in polynomial time. 

I  will  give  you  two  arguments  to  clarify  the  relation  between
$\mathsf{CSP}$ and clauses. Notice that all that follows assumes $\mathbf{P}\not=\mathbf{NP}$. 

First: constraints have a fixed number  of variables,  while  the encoding  of intermediate problems may need  large clauses. This is not  necessarily an issue when such large constraints can  be expressed as a conjunction  of small ones  using auxiliary  variables.  Unfortunately this is not always the case for general $\Gamma$. 

Assume $\Gamma$ to just contain the $\mathsf{OR}$ of five variables.  Clearly you  can express the $\mathsf{OR}$ of less variables   by  repeating   inputs.  You   cannot  express   a  larger $\mathsf{OR}$  because the  way  to do  it  using extension  variables requires  disjunctions  of positive  and  negative literals.  $\Gamma$ represents relations  on variables,  not on literals.  Indeed when you think  about 3-$\mathbf{SAT}$ as  a $\mathsf{CSP}$ you  need $\Gamma$ to contain four relations of disjunction with some negated inputs (from zero to three).

Second: each relation in $\Gamma$ can be expressed as a batch of clauses with (say) three literals.  Each constraint must be a whole batch of such   clauses.  In   the  example   with equality/inequality   constraints    you   cannot   have    a   binary $\mathsf{AND}$ (i.e. relation ${(1,1)}$) without enforcing a binary negated  $\mathsf{OR}$   (i.e.  relation  ${(0,0)}$)   on  the  same
variables.

I hope this illustrates to you that $\mathbf{SAT}$ instances obtained from $\mathsf{CSP}$s have a very peculiar structure, which is enforced by the nature of $\Gamma$.  If the structure is too tight then you cannot express hard problems.  

A corollary of Schaefer Theorem is that whenever $\Gamma$ enforces a structure loose enough to express $\mathbf{NP\backslash P}$ decision problems, then the same $\Gamma$ allows enough freedom to express general 3-$\mathsf{SAT}$ instances.