OK - so your original equation is $$Ax = b$$
Say you've come up with a good preconditioner for $A$, call this $M$. Also, say you have pre-computed an LU-decomposition for this $M$, i.e.
$$M = L_m U_m$$
where I've used the subscript $m$ to indicate that it is the LU decomposition of $M$ that we're talking about here (and not that of $A$). Then, assuming left preconditioning, you will be solving the preconditioned linear system
$$ M^{-1} A x = M^{-1}b,$$
or, equivalently
$$ (L_m U_m)^{-1} A x = (L_m U_m)^{-1} b,$$
or
$$ U_m^{-1} L_m^{-1} A x = U_m^{-1} L_m^{-1} b.$$
Thus the preconditioned system matrix is now $U_m^{-1} L_m^{-1}A$, which is (if you have chosen a good preconditioner) much better conditioned than $A$.
Now, you can
(a) (as a purely academic exercise,) solve the system with a non-preconditioned Krylov method (say GMRES). Because the system matrix is well conditioned, you can expect quick convergence.
The Krylov algorithm will need to multiply this matrix to an arbitrary (known) vector, call that $r$. In other words, what the Krylov procedure needs from you is a routine that, given $r$, computes $$z = U_m^{-1} L_m^{-1} A r$$
This is equivalent to solving the linear system $L_m U_m z = A r$ (for $z$). This system is easy (=cheap) to solve because the matrices $L_m$ and $U_m$ are triangular.
or
(b) (what you'd do in practice) run preconditioned CG on the preconditioned system. CG will apply the inverse preconditioner to the residual vector $r := b - Ax$. Again, this just means it solves the linear system $L_m U_m z = r$ (for $z$).
So, to summarize: never invert matrices explicitly (you can't do that in practice for large matrices anyway), instead solve the corresponding linear system (which is mathematically equivalent to applying the inverse matrix operator to a known vector).