Let $X$ be any infinite set and $f:X^n\to X$ be an $n$-ary operation for some $n>2$. Let $g:X\times X\to X$ be any bijection, and let $h:X^{n-1}\to X$ be given by $h(x_1,x_2,\dots,x_{n-1})=f(y_0,y_1,x_2,\dots,x_{n-1})$ where $y_0,y_1\in X$ are the unique elements such that $g(y_0,y_1)=x_1$. We then have $$f(x_0,x_1,\dots,x_{n-1})=h(g(x_0,x_1),x_2,\dots,x_{n-1}).$$ That is, $f$ is a composition of one binary operation and one $(n-1)$-ary operation. By induction on $n$, this shows that any $n$-ary operation on $X$ can be written as a composition of $n-1$ binary operations (and in fact, all but the last of these binary operations can be chosen to be some fixed bijection $X\times X\to X$).
Note that clearly you can't use fewer than $n-1$ binary operations, since in any expression formed by composing $k$ binary operations, you have only $k+1$ different inputs. So if you had fewer than $n-1$ operations, one of the variables would need to not appear as an input at all, and so the $n$-ary operation could not depend on that variable.
If you impose additional restrictions on what kinds of operations you're allowed to use, this question can become much more subtle, and is generally known as (variations on) Hilbert's thirteenth problem. For instance, in the case $X=\mathbb{R}$, if you require all the operations to be continuous, then it is a theorem of Arnold and Kolmogorov that every operation can be written as a composition of binary operations (or in fact, a composition of addition and unary operations). If you require the operations to be smooth instead, then for every $n$ there are $n$-ary operations on $\mathbb{R}$ which are not compositions of operations of lower arity (see https://mathoverflow.net/questions/195380/are-all-smooth-functions-composites-of-0-1-and-2-ary-functions).
Every operation is a type of function. Not every function is a type of operation. An "operator" is a type of operation, hence a type of function. But not every operation is an operator.
If $A$ and $B$ are sets, a function $f$ from $A$ to $B$ is a subset of $A\times B$ that satisfies the following two conditions:
- For each $a\in A$ there exists $b\in B$ such that $(a,b)\in f$; and
- For each $a\in A$, if $b,b'\in B$ are such that $(a,b)\in f$ and $(a,b')\in f$, then $b=b'$.
If $f$ is a function, and $(a,b)\in f$, we denote this by writing $f(a)=b$. If $f$ is a function from $A$ to $B$, we denote this by writing $f\colon A\to B$.
If $A$ is a set and $n$ is an ordinal (think finite number, but you can do it more generally), then an $n$-ary operation on $A$ is a function $f\colon A^n\to A$. Note in particular that for a function to be an operation, there must exist a set $A$ such that the domain of $f$ is of the form $A^n$ for some ordinal $n$, and the codomain of $f$ is the set $A$.
An operator is a $1$-ary operation, that is a function $f\colon A\to A$.
However, that type of nomenclature is usually reserved for specific circumstances, such as a linear function from a vector space to itself rather than in general.
The function mapping $\mathbb{Q}$ to $\mathbb{R}$ by sending $a\in\mathbb{Q}$ to $a\sqrt{2}\in \mathbb{R}$ is a function, but not an operation.
The function mapping $\mathbb{R}^2\to\mathbb{R}$ by sending $(a,b)$ to $a+b$ is a (binary) operation on $\mathbb{R}$, but not an operator on $\mathbb{R}$.
The function $T\colon \mathbb{R}\to\mathbb{R}$ given by $T(x)=2x+1$ is a (unary) operation on $\mathbb{R}$. It is also a function from $\mathbb{R}$ to itself; it is also an operator on $\mathbb{R}$, though we don't usually refer to it that way.
Best Answer
While your reasoning is correct that "every function is an operation" under that extremely general definition of "operation", I would say that a more common definition of an "operation" on a set $S$ would be a function $$\alpha: S^n\to S\quad\text{ for some }n\geq 0$$ or, to allow "partial" operations, $$\alpha: X\to S\quad\text{ where }X\subset S^n\text{ for some }n\geq 0$$ (and we would say $\alpha$ is an $n$-ary operation). Under this definition, there are clearly many functions that are not operations.