The problem is unclear as presently written, because you are comparing a set of population values $(x_i,y_i)$ with a pair of random variables $(X,Y)$ where the latter have not been clearly defined. However, suppose we let $I \sim \text{U} \{ 1,...,N \}$ denote a random value from the population and we then define the values:

$$X \equiv x_I \quad \quad \quad Y \equiv y_I.$$

This definition means that $(X,Y)$ is now a random pair of points from the finite population. With a bit of work, we can now show the equivalence between these two formulae. As a preliminary result we can use the law of iterated expectation to get:

$$\begin{align} \mu_X \equiv \mathbb{E}(X) &= \mathbb{E}(\mathbb{E}(X|I)) \\[12pt] &= \mathbb{E}(X_I) \\[6pt] &= \frac{1}{N} \sum_{i=1}^N x_i \\[6pt] &= \bar{x}_N, \\[18pt] \mu_Y \equiv \mathbb{E}(Y) &= \mathbb{E}(\mathbb{E}(Y|I)) \\[12pt] &= \mathbb{E}(Y_I) \\[6pt] &= \frac{1}{N} \sum_{i=1}^N y_i \\[6pt] &= \bar{y}_N. \\[12pt] \end{align}$$

Another application of the law of iterated expectation then gives:

$$\begin{align} \mathbb{E}((X-\mu_X)(Y-\mu_Y)) &= \mathbb{E}(\mathbb{E}((X-\mu_X)(Y-\mu_Y)|I)) \\[16pt] &= \mathbb{E}((X_I-\mu_X)(Y_I-\mu_Y)) \\[8pt] &= \frac{1}{N} \sum_{i=1}^N (x_i-\mu_X)(y_i-\mu_Y) \\[6pt] &= \frac{1}{N} \sum_{i=1}^N (x_i-\bar{x}_N)(y_i-\bar{y}_N). \\[6pt] \end{align}$$

So, as you can see, these two formulae are equivalent if you define the pair $(X,Y)$ to be a random pair in the population.

The problem is unclear as presently written, because you are comparing a set of population values $(x_i,y_i)$ with a pair of random variables $(X,Y)$ where the latter have not been clearly defined. However, suppose we let $I \sim \text{U} \{ 1,...,N \}$ denote a random value from the population and we then define the values:

$$X \equiv x_I \quad \quad \quad Y \equiv y_I.$$

This definition means that $(X,Y)$ is now a random pair of points from the finite population. With a bit of work, we can now show the equivalence between these two formulae. As a preliminary result we can use the law of iterated expectation to get:

$$\begin{align} \mu_X \equiv \mathbb{E}(X) &= \mathbb{E}(\mathbb{E}(X|I)) \\[12pt] &= \mathbb{E}(X_I) \\[6pt] &= \frac{1}{N} \sum_{i=1}^N x_i \\[6pt] &= \bar{x}_N, \\[18pt] \mu_Y \equiv \mathbb{E}(Y) &= \mathbb{E}(\mathbb{E}(Y|I)) \\[12pt] &= \mathbb{E}(Y_I) \\[6pt] &= \frac{1}{N} \sum_{i=1}^N y_i \\[6pt] &= \bar{y}_N. \\[12pt] \end{align}$$

Another application of the law of iterated expectation then gives:

$$\begin{align} \mathbb{E}((X-\mu_X)(Y-\mu_Y)) &= \mathbb{E}(\mathbb{E}((X-\mu_X)(Y-\mu_Y)|I)) \\[16pt] &= \mathbb{E}((X_I-\mu_X)(Y_I-\mu_Y)) \\[8pt] &= \frac{1}{N} \sum_{i=1}^N (x_i-\mu_X)(y_i-\mu_Y) \\[6pt] &= \frac{1}{N} \sum_{i=1}^N (x_i-\bar{x}_N)(y_i-\bar{y}_N). \\[6pt] \end{align}$$

As you can see, these two formulae are equivalent if you define the pair $(X,Y)$ to be a random pair in the population. You needn't worry about "convergence" here, since the two formulae are equivalent for all $1 \leqslant N < \infty$.

This equality is part of the "design-based" approach to sampling theory, where we implicitly condition on the empirical distribution of the population, and take our random variable version of a data value to be a random value from this distribution. Note that things are different in the "model-based" approach, where we would usually define a random observation as coming from a higher-level "superpopulation" (infinite population) in which the finite population is embedded. In the latter case the moment quantities for a random observation usually refer to moments of the superpopulation distribution, rather than the finite population empirical distribution.

The problem is unclear as presently written, because you are comparing a set of population values $(X_i,Y_i)$ with a pair of random variables $(X,Y)$ where the latter have not been clearly defined. However, suppose we let $I \sim \text{U} \{ 1,...,N \}$ denote a random value from the population and we then define the values:

$$X \equiv X_I \quad \quad \quad Y \equiv Y_I.$$

This definition means that the random pair $(X,Y)$ is now a random pair of points from the finite population. With a bit of work, we can now show the equivalence between these two formulae. As a preliminary result we can use the law of iterated expectation to get:

$$\begin{align} \mu_X \equiv \mathbb{E}(X) &= \mathbb{E}(\mathbb{E}(X|I)) \\[12pt] &= \mathbb{E}(X_I) \\[6pt] &= \frac{1}{N} \sum_{i=1}^N x_i \\[6pt] &= \bar{x}_N, \\[18pt] \mu_Y \equiv \mathbb{E}(Y) &= \mathbb{E}(\mathbb{E}(Y|I)) \\[12pt] &= \mathbb{E}(Y_I) \\[6pt] &= \frac{1}{N} \sum_{i=1}^N y_i \\[6pt] &= \bar{y}_N. \\[12pt] \end{align}$$

Another application of the law of iterated expectation then gives:

$$\begin{align} \mathbb{E}((X-\mu_X)(Y-\mu_Y)) &= \mathbb{E}(\mathbb{E}((X-\mu_X)(Y-\mu_Y)|I)) \\[16pt] &= \mathbb{E}((X_I-\mu_X)(Y_I-\mu_Y)) \\[8pt] &= \frac{1}{N} \sum_{i=1}^N (x_i-\mu_X)(y_i-\mu_Y) \\[6pt] &= \frac{1}{N} \sum_{i=1}^N (x_i-\bar{x}_N)(y_i-\bar{y}_N). \\[6pt] \end{align}$$

So, as you can see, these two formulae are equivalent if you define the pair $(X,Y)$ to be a random pair in the population.

The problem is unclear as presently written, because you are comparing a set of population values $(x_i,y_i)$ with a pair of random variables $(X,Y)$ where the latter have not been clearly defined. However, suppose we let $I \sim \text{U} \{ 1,...,N \}$ denote a random value from the population and we then define the values:

$$X \equiv x_I \quad \quad \quad Y \equiv y_I.$$

This definition means that $(X,Y)$ is now a random pair of points from the finite population. With a bit of work, we can now show the equivalence between these two formulae. As a preliminary result we can use the law of iterated expectation to get:

$$\begin{align} \mu_X \equiv \mathbb{E}(X) &= \mathbb{E}(\mathbb{E}(X|I)) \\[12pt] &= \mathbb{E}(X_I) \\[6pt] &= \frac{1}{N} \sum_{i=1}^N x_i \\[6pt] &= \bar{x}_N, \\[18pt] \mu_Y \equiv \mathbb{E}(Y) &= \mathbb{E}(\mathbb{E}(Y|I)) \\[12pt] &= \mathbb{E}(Y_I) \\[6pt] &= \frac{1}{N} \sum_{i=1}^N y_i \\[6pt] &= \bar{y}_N. \\[12pt] \end{align}$$

Another application of the law of iterated expectation then gives:

$$\begin{align} \mathbb{E}((X-\mu_X)(Y-\mu_Y)) &= \mathbb{E}(\mathbb{E}((X-\mu_X)(Y-\mu_Y)|I)) \\[16pt] &= \mathbb{E}((X_I-\mu_X)(Y_I-\mu_Y)) \\[8pt] &= \frac{1}{N} \sum_{i=1}^N (x_i-\mu_X)(y_i-\mu_Y) \\[6pt] &= \frac{1}{N} \sum_{i=1}^N (x_i-\bar{x}_N)(y_i-\bar{y}_N). \\[6pt] \end{align}$$

So, as you can see, these two formulae are equivalent if you define the pair $(X,Y)$ to be a random pair in the population.