The issue is not so much about the use of the term "observations" versus "samples" or "random variables" or "realizations," but rather, a misconception about the meaning of the expectation operator $\operatorname{E}$.
Suppose I have a single random variable, say $X$, that follows some probability distribution. To make things very simple, let's suppose this distribution is
$$\Pr[X = 2] = 0.3, \quad \Pr[X = 5] = 0.7.$$
If I ask you, "what is $\operatorname{E}[X]$," what would you do? Would you insist that you cannot calculate it because you have no sample?
Of course, $$\operatorname{E}[X] = 2 \Pr[X = 2] + 5 \Pr[X = 5] = (2)(0.3) + (5)(0.7) = 4.1.$$ This is not an empirical result; it does not depend on any observation of $X$ whatsoever.
Now suppose I have two of these random variables, say $X_1$ and $X_2$, independent and identically distributed as $X$. What would you say about $\operatorname{E}[X_1 + X_2]$? We could either explicitly compute the probability distribution of the sum $X_1 + X_2$; i.e.,
$$\begin{align}
\Pr[X_1 + X_2 = 4] &= \Pr[X_1 = 2]\Pr[X_2 = 2] = 0.09 \\
\Pr[X_1 + X_2 = 7] &= \Pr[X_1 = 2]\Pr[X_2 = 5] + \Pr[X_1 = 5]\Pr[X_2 = 2] = 0.42 \\
\Pr[X_1 + X_2 = 10] &= \Pr[X_1 = 5]\Pr[X_2 = 5] = 0.49 \\
\end{align}$$
hence
$$\operatorname{E}[X_1 + X_2] = (4)(0.09) + (7)(0.42) + (10)(0.49) = 8.2,$$
or we can use linearity of expectation:
$$\operatorname{E}[X_1 + X_2] = \operatorname{E}[X_1] + \operatorname{E}[X_2] = 4.1 + 4.1 = 8.2.$$
Thus the expectation of a sample mean of size $n = 2$ is
$$\operatorname{E}\left[\frac{X_1 + X_2}{2}\right] = \frac{1}{2}\operatorname{E}[X_1 + X_2] = 4.1.$$
And in the general case of an arbitrary sample size, the same idea holds. The expectation operator is not calculating a sample mean. It's calculating a distributional property, a measure of central tendency of the underlying distribution. If I write
$$\operatorname{E}[\bar X] = \operatorname{E}\left[\frac{X_1 + \cdots + X_n}{n}\right],$$
this represents a distributional property of the sampling distribution of the sample mean. That is to say, just like how $X_1 + X_2$ has its own probability distribution (that we can derive from $X$), the sample mean $\bar X$ has a probability distribution. $\operatorname{E}[\bar X]$ is the expected value of that distribution.
An "observation" for which we have no numerical result remains a random variable for the purposes of calculating distributional properties, except where noted otherwise. It doesn't matter if I call $X_1$ an "observation" or a random variable if I want to compute $\operatorname{E}[X_1 + X_2 + \cdots + X_n]$. The context is clear: I want an expectation with respect to all of the variables $X_1, \ldots, X_n$. If I want to treat $X_1$ as a fixed but unknown value, I would write
$$\operatorname{E}[X_1 + \cdots + X_n \mid X_1],$$
and this would be a function of $X_1$. But in no case should you be thinking about $\operatorname{E}$ as being some empirical mean of some hypothetical sample drawn from a probability distribution. $\operatorname{E}[X]$ is not a statistic and not a random variable.
It is strange, because this is the second recent instance I have seen where someone is confused between estimators and distributional properties (i.e., parameters). This should not be the case for students of probability and statistics; it is critically important to disambiguate these before proceeding any further.
