I'm working on a problem (namely, if $\alpha + \beta$ is algebraic over $F$ then $\alpha$ is algebraic over $F[\beta]$), and the binomial formula appeared.
For the problem, I used the fact that, for the polynomial in $F[x]$ with $\alpha + \beta$ as a root, $$\sum^{k}_{i=0}c_i (\alpha + \beta)^i = 0$$ $$= \sum^{k}_{i=0}\dbinom{k}{i}c_i\alpha^i \beta^{k-i}$$
I'm hoping to say that $\dbinom{k}{i}c_i\beta^{k-i} \in F[\beta]$ but I'm not sure how to think of the binomial coefficient here - should it be considered as $\dbinom{k}{i}$ copies of the "+" operation of $F$? I know that the binomial formula holds for an arbitrary field, but I don't feel confident enough to say that these coefficients do belong to $F$. It feels unnatural multiplying an element from $F$ by some natural number.
I may be missing something obvious here, but any clarification is appreciated.
Your interpretation of the binomial coefficient as repeated addition (of $1$'s) is correct. Moreover, you should not feel that it is in any way unnatural to multiply an element of $F$ by an integer in $\mathbb{Z}$. Indeed, the additive group of $F$ is an abelian group, i.e. a $\mathbb{Z}$-module, and the notation $na = n \cdot a = a + \ldots + a$ represents the action of the integer $n$ on the field element $a$. However, keep in mind that if the characteristic is not $0$, then equalities such as $p = 0$ will hold in the field.
The same is true for an arbitrary ring, and indeed the binomial formula holds over any commutative ring.