In David Tong's lecture notes on statistical field theory, he considers the partition function of the Ising model and computes the effective free energy by integrating out the microscopic details of individual configurations. Why is it suddenly free energy? The Boltzmann factor always has inverse temperature and energy in its exponent. The appearance of free energy does not make sense to me here.
It is the definition of free energy in the canonical ensemble. For the entire system: $$ e^{-\beta F} = \sum_{i\in I} e^{-\beta E_i} $$ where you sum over all the microstates indexed by $I$. To compute the free energy of a subset $A$, just restrict the sum: $$ e^{-\beta F(A)} = \sum_{i\in A} e^{-\beta E_i} $$ This allows the coarse grained approach. Let $(A_j)_{j\in J}$ be a partition of the microstates, then you exactly get: $$ e^{-\beta F} = \sum_{j\in J} e^{-\beta F(A_j)} $$ Note that this generalises the original definition since the microstate energy is just its free energy as well since the entropy is zero. Intuitively, you recover free energy because there is both an energy and entropy contribution to the sum. The entropy is due to the multiplicity of the terms and energy is due to the magnitude of the terms.
Perhaps you are confused because you are more used to the microcanonical ensemble. You can check the usual equivalence in the thermodynamic limit, i.e. for macrostates. Let $W(E)$ be the number of microstates of energy $E$, then the usual microcanonical definitions are: $$ W(E) = e^{S_m(E)} \\ \beta = \partial_ES \\ \beta F_m(E) = \beta E-S $$ You can therefore rewrite the sum from the canonical ensemble as: $$ \begin{align} e^{\beta F_c} &= \sum_E e^{S_m-\beta E} \\ &\asymp e^{\beta F_m(E_c)} \end{align} $$ The logarithmic leading order is given by Laplace's method, where $E_c$ is where the argument of the exponential is maximised: $$ E_c = \text{argmax}_E S_m(E)-\beta E $$ which therefore satisfies: $$ \partial_ES_m(E_c)-\beta = 0 $$ You therefore recover the equivalence: $$ F_c(\beta) = \inf_E \beta E-S_m(E) \\ F_c(\beta) = F_m(E_c) $$ i.e. $F_c$ is the Legendre transform of $S_m$ as you might have learned in thermodynamics.
Typically for field theories, which are mesoscopic theories, you coarse grain over macrostates (physically, you apply block spins a macroscopic number of times), so the previous analysis holds.
