I think the notes you refer to are unclear, and they may easily confuse even an expert reader.
The argument starts with an example valid for a system at equilibrium at a constant temperature, volume, and number of particles.
By using the standard definition of microstate (the full specification of all the microscopic degrees of freedom) and using the usual meaning of the canonical ensemble probability distribution for the microstates, the author introduces the probability distribution for the energy values, which implies a factor $\Omega(E_A)$, corresponding to the number of microstates with the same energy $E_A$. At this point, $\Omega$ is rewritten in terms of the entropy of a microcanonical ensemble of fixed energy $E_A$.
The final step, which would require some justification, is to conclude that the probability of a macrostate at fixed energy $E_A$ is given by $e^{-\beta F}/Z$, where $F = E_A - ST$ is identified with the Helmholtz free energy.
Such identification is possible only at the thermodynamic limit. The equivalence of different ensembles allows us to identify the entropy $S$ of the previous formula, the microcanonical entropy, with the entropy of the canonical ensemble.
Once this result has been obtained, the relation between $F$ and $Z$ is left untouched. The result is just a way to use the canonical probability of the microstates to derive the probability of finding a system at fixed $T$, $V$, and $N$, at a given value of the macroscopic energy.