That's correct, $X_1, X_2, ..., X_n$ are $n$ independent and identically distributed (i.i.d.) values of the same variable drawn from the same distribution $f(x)$.
The term "i.i.d." means the values of $X$ are completely independent (the probability that $X$ assumes a particular value for record $i$ is unrelated to the values of $X$ for other records) and all values of $X$ are random pulls from the same distribution (e.g., the normal distribution with a given mean and variance).
When I was an undergraduate, the professor in my probability class began each lecture by drawing two balls in succession (without replacement) from an urn that he brought to class. Some days, the first ball was white and the second ball black, while on other days, the first ball was black and the second ball was white. I noticed over the course of the semester that roughly half the time, the first ball was white and the second black, and half the time it was the other way around. So, I figured that the the probability that the first ball was white was $0.5$ and the probability that the second ball was white was also $0.5$.
A classmate of mine was always just a tad late coming to class and he observed only the second ball being drawn and he also noted that roughly half the time, the ball that our professor drew was white, and he too estimated the probability that the professor drew a white ball was $0.5$. He didn't know that the ball that our profossor was drawing as my friend walked in was the second ball that the professor was drawing from the urn. And yet, my friend and I came up with the same estimate of the probability of the (second) ball being white.
At the end of the semester, our professor invited the class to examine the urn. I was surprised to discover that the urn contained only one black ball and one white ball! That explained why the draws were always (white, black) or (black, white). By golly, those draws were dependent as heck but they both had the same marginal probability $0.5$ of resulting in a white ball both for me who saw both draws and for my classmate who didn't know that he was observing the result of the second draw from the urn.
More generally, in sampling without replacement from a population of $n$ distinct items, suppose that we are taking $k < n$ samples. Then the $k$ samples are all distinct. Unknown to us, God continues sampling without replacement until all $n$ items have been draw. God's experiment has $n!$ different outcomes each of which has probability $\dfrac{1}{n!}$. How many of these outcomes have item #i occurring in the $j$-th place? Well, God's experiment has $n!$ possible outcomes of which exactly $(n-1)!$ outcomes have item #i in the $j$-th place (and the $n-1$ outcomes #1, #2, $\ldots$, #(i-1), #(i+1), #(i+2), $\ldots$, #n scattered about in places $1, 2, \ldots, (i-1), (i+1), \ldots n$. So, at least in God's mind, the probability that item #i occurs in the $j$-th place is $\dfrac{(n-1)!}{n!} = \dfrac 1n$ regardless of what $j$ is. In God's mind, item #i has the same probability $\dfrac 1n$ of occurring in each of the $n$ places. To the extent that we all hope to know what is in God's mind, we should accept these calculations as correct, even though we stopped after $k$ draws and didn't complete the experiment by drawing all $n$ items and so didn't get to see what God obtained in draws numbered $k+1, k+2, \cdots, n$.
Note that the events that "item #i occurs in the $j$-th place"and "item #i occurs in the $j^\prime$-th place" are disjoint events (the cannot occur simultaneously), not independent events. Very dependent but nonetheless equally likely
Best Answer
A dataset is a sample of the population. A dataset is not a random variable, which has a precise mathematical definition, i.e. it is a (measurable) function (you can ignore the "measurable" adjective!): you can simply think of a random variable as a map between outcomes of e.g. an experiment and real numbers.
A dataset (or a sample) contains $N$ "realisations" (or "outcomes") of one or more random variables, where $N$ is the size of the dataset. More precisely, each value associated with each feature of the dataset is (usually) or can be associated with one random variable: when you sample from the population (i.e., you get one row of the dataset), each of the random variables is "realized", that is, you obtain one of the concrete outcomes that the associated random variable can take.
A dataset can thus be considered the realization(s) of a random variable, if there is only one feature, or of multiple random variables (or, equivalently, of a multi-dimensional or multi-variate random variable), if there are multiple features.
In your example, if $X=(x_1, x_2, \dots, x_M)$ is one row of your dataset, then $X$ can be associated with the "realization" of $M$ random variables: $x_1, x_2, \dots, x_M$ are the realizations of these $M$ random varables (which we can denote by $X_1, X_2, \dots, X_M)$, that is, they are the outcomes of these random variables. You may have more than one row or, equivalently, maybe $X$ is actually a matrix, where $x_i$ is a column vector of size $N$ (i.e. the size of the dataset) which contains $N$ realizations of the random variable associated with $x_i$ (which can be denoted by $X_i$).
In conclusion, the term "feature" is (usually) equivalent to the term "random variable". A random variable (or feature) can have several realisations (or outcomes).