Is the set of all differentiable functions $f: \Re \rightarrow \Re$ such that $f'(0) = 0$ is a vector space over $\Re$? I was given the answer yes by someone who is better at math than me and he tried to explain it to me, but I don't understand. I am having difficulty trying to conceptualize this idea of vector spaces with functions because I can't really visualize it like a plane in 3d space. I am also wondering what is the importance of having vector spaces set over a field? It seems trivial or maybe its just me being brainwashed by years of elementary mathematics
A vector space is merely a set with two operations, addition and scalar multiplication, that satisfy certain conditions. In this case the scalars are real numbers. The addition operation is the pointwise sum, and scalar multiplication is multiplication by a real number.
Besides these properties it geometrically has little to do with vectors in three dimensional space, and the concept of a vector having "both a magnitude and direction" no longer makes much sense (without additional structure). This is the definition of a vector space that has been settled on, so getting used to it would be a good idea.
That said, vector spaces are incredibly well behaved and a great deal of algebraic material from the finite dimensional case generalizes to infinite dimensional vector spaces without modification. This is part of the utility of vector spaces; once you have a set that satisfies a few easy axioms you can do linear algebra in it.
To answer the second question: The importance of the vector space being over a field is that having a field makes many things easier. There exists "vector spaces" (called modules) over non-fields (for example, over the integers), but a lot of things we rely on when working with vector spaces don't necessarily hold for them. For example, a module might not even have a basis.
So basically we require the vector space to be over a field because that gives us useful properties.
The space P^3 of polynomials of the type a_1x+a_2x^2+a_3x^3 is the same as R^3 and my be regarded as a 3D Euclidean space.