If the map from $X$ to the maxspec of $\mathcal O(X)$ is a bijection, then $X$ is indeed affine.
Here is an argument:
By assumption $X \to $ maxspec $\mathcal O(X)$ is bijective, thus quasi-finite,
and so by (Grothendieck's form of) Zariski's main theorem, this map factors as an open embedding of $X$ into a variety that is finite over maxspec $\mathcal O(X)$.
Any variety finite over an affine variety is again affine, and hence $X$ is an open subset of an affine variety, i.e. quasi-affine. So we are reduced to considering the case when $X$ is quasi-affine, which is well-known and straightforward.
(I'm not sure that the full strength of ZMT is needed, but it is a natural tool
to exploit to get mileage out of the assumption of a morphism having finite fibres, which is what your bijectivity hypothesis gives.)
In fact, the argument shows something stronger: suppose that we just assume
that the morphism $X \to $ maxspec $\mathcal O(X)$ has finite non-empty fibres,
i.e. is quasi-finite and surjective.
Then the same argument with ZMT shows that $X$ is quasi-affine. But it is standard that the map $X \to $ maxspec $\mathcal O(X)$ is an open immersion when $X$ is quasi-affine,
and since by assumption it is surjecive, it is an isomorphism.
Note that if we omit one of the hypotheses of surjectivity or quasi-finiteness, we can find a non-affine $X$ satisfying the other hypothesis.
E.g. if $X = \mathbb A^2 \setminus \{0\}$ (the basic example of a quasi-affine,
but non-affine, variety), then maxspec $\mathcal O(X) = \mathbb A^2$, and the open immersion $X \to \mathbb A^2$ is evidently not surjective.
E.g. if $X = \mathbb A^2$ blown up at $0$, then maxspec $\mathcal O(X) =
\mathbb A^2$, and $X \to \mathbb A^2$ is surjective, but has an infinite fibre
over $0$.
Caveat/correction: I should add the following caveat, namely that it is not always true, for a variety $X$ over a field $k$, that $\mathcal O(X)$ is finitely generated over $k$, in which case maxspec may not be such a good construction to apply, and the above argument may not go through. So in order to conclude that $X$ is affine, one should first insist that $\mathcal O(X)$ is finitely generated over $k$, and then that futhermore the natural map $X \to $ maxspec $\mathcal O(X)$ is quasi-finite and surjective.
(Of course, one could work more generally with arbitrary schemes and Spec rather than
maxspec, but I haven't thought about this general setting: in particular, ZMT requires some finiteness hypotheses, and I haven't thought about what conditions might guarantee that the map $X \to $ Spec $\mathcal O(X)$ satisfies them.)
Incidentally, for an example of a quasi-projective variety with non-finitely generated ring of regular functions, see this note of Ravi Vakil's
This is just a matter of terminology. In both books I have to hand (Hartshorne, and Eisenbud's "Commutative algebra..."), the authors define an 'affine algebraic set' to be any subset of $\mathbb{A}^n$ given by the vanishing of polynomials, and an 'affine algebraic variety' to be an irreducible such set.
What is perfectly clear (and is possibly what 'The question' really asks, given the absence of the word 'variety' at the end) is that any closed subset of an affine algebraic set is again an affine algebraic set.
In practice though, the word 'variety' often seems to be used more generally, without the irreducibility requirement.
Best Answer
The answer to your question is that coordinate algebras are intrinsic - they don't depend upon the embedding of our variety in to a bigger space. Therefore it doesn't matter where our variety is embedded: every way to compute the coordinate algebra will give us isomorphic answers. So your final line about viewing $f$ as a function on $W$ via the pullback is fine. (As you spend more time in algebraic geometry, this potential gap in your understanding should disappear for definitional reasons: the objects that track what "functions" are on a variety or their generalizations become part of the data of the object, and they always play nice with embeddings.)