Is the set of all functions $f: R → R$ such that $f(0) = 0$ a vector space? Is this true? Also, I need some help wrapping my head around the concept...how is a function expressed as a vector space?
 A: Hint: You need to start with the end of your question:

how is a function expressed as a vector space?

It's not. A set of real valued functions of a real variable may be a vector space. To see that, think about how you would add two functions, and how you would multiply a function by a scalar (two things you did routinely in calculus). Then look at the vector space axioms.
Once this is clear to you the first part should be straightforward.
A: A vector space isn't generally defined in the way we typically think of them as a list of coordinates we add component-wise. In general, we think of a vector space as a family of objects (called vectors) which we can add together and multiply by real numbers, where the addition and multiplication "play well" together (you can look up the vector space axioms).
What this means is we often talk about vector spaces where we can't really think of the vectors as lists of coordinates, but instead must take the addition as it's presented to us. But oftentimes, we instead define these operations in other familiar ways. For instance, we might add together sequences by adding each term. Or we might talk about adding polynomials, or even more general families of functions.
So don't think of it as "expressing" these functions as vector spaces. The claim is that the functions can be added together and multiplied by real numbers in such a way that the family is a vector space. So you should go through the vector space axioms and determine whether the family in question satisfies the axioms, or whether instead some are violated.
A: Wrapping your head around the concept.
Suppose we look at the set of functions that are defined at $1,2,3.$
Then we could define the vector $\mathbf f =  <f(1),f(2), f(3)>$ and wouldn't this behave like any vector in $\mathbb R^3$ you have worked with?
You could add it to another function using the same rules, and you can multiply it by a scalar.
We could extend this to an arbitrary number of integers and you have a vector in $\mathbb R^n.$
So, is it really such a leap to extend the concept to functions defined over the real numbers?
A: Specifically, the set of all functions, f, R->R, such that f(0)= 0, forms a vector space since:
1) The 0 vector, f(x)= 0 for all x, is in this set.
2) If f and g are two functions in this set (so f(0)= 0 and g(0)= 0), then their sum, f+ g, has the property that (f+ g)(0)= f(0)+ g(0)= 0+ 0= 0.
3) If f is a function in this set (so f(0)= 0), and a is any real number, then the product, af, has the property that (af)(0)= af(0)= a(0)= 0.
(Note that instead of "1) the 0 vector is in this set." we could just say "1)this set is non-empty". Once we have at least one vector, v, in the set, by (3), -1v is also in the set and then, by (2), v+ (-1)v= 0 is in the set.)
