I think I may have a solution for this one, but I want to share to validate if anyone else reaches a similar way to solve the problem.

First, we start with the definition of the expectation of the Average Test Error:

$$\mathbb{E}_{p(\vec{x}, y)}[\text{Average Test Error}]$$

$$=\mathbb{E}_{p(\vec{x}, y)}[\frac{1}{M}\sum_{m=1}^M L(y^{(m)}, f(\vec{x}^{(m)}))] \quad \vec{x}^{(m)}, y^{(m)}\sim p(\vec{x}, y) $$

Since the expectation is a linear operator as seen in this note, we move the expectation operator inside the summation:

$$\frac{1}{M}\sum_{m=1}^M \mathbb{E}_{p(\vec{x}, y)}[L(y^{(m)}, f(\vec{x}^{(m)}))] \quad \vec{x}^{(m)}, y^{(m)}\sim p(\vec{x}, y)$$

From the definition of expectation, we have: $$\mathbb{E}{p(\vec{x}, y)}[L(y^{(m)}, f(\vec{x}^{(m)}))] \quad \vec{x}^{(m)}, y^{(m)}\sim p(\vec{x}, y)$$ $$=\int\limits{\vec{x}}\int\limits_{y}L(y, f(\vec{x}))p(\vec{x},y) \, dy \, d\vec{x}$$

Therefore our expectation becomes:

$$\frac{1}{M}\sum_{m=1}^M \int\limits_{\vec{x}}\int\limits_{y}L(y, f(\vec{x}))p(\vec{x},y) \, dy \, d\vec{x}$$

Which is simply an average over a constant, thus we have:

$$\mathbb{E}_{p(\vec{x}, y)}[\text{Average Test Error}]$$

$$ = \int\limits_{\vec{x}}\int\limits_{y}L(y, f(\vec{x}))p(\vec{x},y) \, dy \, d\vec{x}$$

Which is our generalization error.