Let $K$ be a field. A vector space over $K$ is a set $V$, together with an operation $+\colon V\times V\to V$, and a function $\cdot\colon K\times V\to V$ (or alternatively, a family of functions $\lambda_c\colon V\to V$, indexed by elements $c\in K$) subject to certain identities and conditions.
The set $V$ does not normally contain $K$, so it makes no sense to talk about the vector space $V$ as being "closed under scalar multiplication". In addition, there is in general no "product" of vectors, so it does not make sense to talk about "vector multiplication."
Even in the case where we have a cross product of vectors (e.g., $\mathbb{R}^3$), this operation does not make $V$ into a group (the product is not associative).
Scalar multiplication is not an operation; an operation is always a function from a cartesian power of a set to the set. However, you can use currying to view the scalar multiplication as a family of unary operations on $V$, as indicated above, where for each $c\in K$ and each $\mathbf{v}\in V$, we define $\lambda_c(\mathbf{v})=c\mathbf{v}$.
(Added. Once you take into account the vector space axioms that relate to the scalar multiplication, it turns out that the scalar multiplication $K\times V\to V$ is actually an action of the field $K$ on the additive group $(V,+)$. But to truly call it an "action", you need the function to satisfy certain properties; just having a function $K\times V\to V$ does not make it an action, whereas any function $S\times S\to S$ is a binary operation on $S$, regardless of its properties.)
So pretty much all of your questions are based on misstatements of fact. Seems hard to answer them accurately.
Now, there is a situation in which some of what you say might make sense: if $F$ and $K$ are fields, and $F\subseteq K$, then we can view $K$ as a vector space over $F$ by "forgetting" about multiplication inside of $K$ when neither factor is in $F$. But here you really are going the other way: you already have a field, and you are obtaining a vector space by restricting the multiplication $\cdot\colon K\times K\to K$ to $F\times K$.
There is another situation in which you may have a product between "vectors": when you have an algebra. If $F$ is a field, a $K$ algebra is a ring with (a copy of) $F$ in the center of $K$. But again, what you have is a richer structure that, by "forgetting" part of the structure, yields a vector space. Much like you can have a ring and, by forgetting the product, obtain a group.
You can't use the set of functions as scalars since they don't form a field. Functions that have zeroes somewhere don't have inverses. This has nothing to do with periodicity.
Restricting the scalars to functions that are never $0$ won't fix things. The sum of two such functions can have a root, so the set of those functions aren't closed under addition.
Your last PS is confusing, but hints at using the field of integers modulo a prime $p$ for the scalars. That will work: see the wikipedia entry on
finite fields
.
Best Answer
In writing this question I just found the answer -- the key is to remember that the term "extension field" just means a field containing the original field, i.e. the "opposite" of a subfield. There is a lot more written about extension fields than subfields (because of Galois theory), so when one phrases the question in terms of extension fields rather than subfields, one finds the answer quickly.