Whether or not we have a vector space depends on how you interpret "quadratic function whose graph passes through the origin."
In order for us to have a vector space, we will need, for example, to think of $3x$ as a quadratic function. For certainly $x^2+x$ is a quadratic function that passes through the origin, as is $-x^2+2x$. So if we are to have closure under addition, the function $(x^2+x)+(-x^2+2x)$, that is, $3x$, will have to be in the collection.
Indeed the identically $0$ function has to be in our collection, for two reasons.
If we do not allow it, then closure under addition can fail, since $(x^2-x)+(-x^2+x)=0$. Closure under multiplication by the constant $0$ also fails.
It is not difficult, however, to show that the set of functions of the form $ax^2+bx$, where $a$ and $b$ range over all the reals, is a vector space over the reals. Quite a number of axioms need to be verified, but each verification is easy.
As to your question about the "passing through $0$" part, one can also show that the set of all polynomial functions of degree $\le 2$ also is a vector space over the reals. It just is a different vector space than the one under consideration.
Remark: $1.$ In answering the homework question, it may be useful to be cautious. I would suggest doing the following: (i) Observe that if we interpret "quadratic" as meaning that the coefficient of $x^2$ is non-zero, then we do not have a vector space. (Of course one should explain why.) (ii) Show in detail that the collection of all functions of the shape $ax^2+bx$ is a vector space. Many of the steps can be possibly omitted. The key properties that have to be verified are closure under addition and under multiplication by scalars.
$2.$ You asked how to describe the vectors. Might as well use the standard polynomial notation, as in the answer above. The "pass through the origin" part just means the constant term is $0$.
Best Answer
Here's an example. Let $V$ be the set of all $n$-tuples of strictly positive numbers $x_1,\ldots,x_n$ satisfying $x_1+\cdots+x_n=1$. Define "addition" of such vectors by
$$ (x_1,\ldots,x_n) \mathbin{\text{“}{+}\text{''}} (y_1,\ldots,y_n) = \frac{(x_1 y_1,\ldots,x_n y_n)}{x_1 y_1 + \cdots + x_n y_n }. $$
This is a vector space whose zero element is $$ \left( \frac 1 n , \ldots, \frac 1 n \right). $$ The additive inverse of $(x_1,\ldots,x_n)$ is $$ \frac{\left( \dfrac 1 {x_1}, \ldots, \dfrac 1 {x_n} \right)}{\dfrac 1 {x_a} + \cdots + \dfrac 1 {x_n}}. $$ This operation is involved in a basic identity on conditional probabilities: $$ (\Pr(A_1),\ldots,\Pr(A_n)) \mathbin{\text{“}{+}\text{''}} k\cdot(\Pr(D\mid A_1),\ldots,\Pr(D\mid A_n)) = (\Pr(A_1\mid D),\ldots,\Pr(A_n\mid D)) $$ where $k$ is whatever it takes to make the sum of the entries $1$. However, in practice, one wouldn't bother with $k$; just multiply term by term and then normalize.
Here's a more down-to-earth example. Look at $\mathbb R^3$ and say you want to put the zero point at $\vec p = (2,3,7)$. Then define "addition" as follows: $$ \vec a \mathbin{\text{“}{+}\text{''}} \vec b = \underbrace{\vec p + (\vec a - \vec p) + (\vec b - \vec p)}_{\begin{smallmatrix} \text{These are the usual} \\ \text{addition and subtraction.} \end{smallmatrix}}. $$