With practice, one learns to recognize the sort of things that may go wrong with potential "vector spaces", and quickly zoom in on those. But, the thing is, it takes practice to figure this out.
Often, if one thing goes wrong, lots of things will go wrong; sometimes, it is one and only one thing that goes wrong (and it may be hard to spot). At this stage, it might actually be a good idea for you to check each axiom and see whether it is met or not met, because it will afford you a lot of practice. Even though it's enough to find one axiom that fails for something to not be a vector space, finding all the ways in which things go wrong is likely good practice at this stage.
For example, you don't say which problem "says the answer is Axiom 4", and in fact I see no problem, among the ones listed, in which $4x+1$ is even a vector! It's not a $4\times 6$ matrix, it's not a $1\times 1$ matrix, it's not a degree 3 polynomial, it's not a degree 5 polynomial, it's not a first degree polynomial whose graph passes through the origin, and it's not a quadratic function whose graph passes through the origin...
Since user6312 already got you started with Question 15, let's continue: you know it fails Axiom 1. It is not hard to verify that it satisfies axioms 2 and 3. Axiom 4 fails because the zero vector (the polynomial 0) is not in your set....
Axiom 5 is a bit tricky: strictly speaking, Axiom 5 does not even make sense if Axiom 4 fails, because there is no $\mathbf{0}$ in the first place. I would certainly score such a statement as correct. On the other hand, if you have a polynomial of degree exactly 3, $ax^3+bx^2+cx+d$, with $a\neq 0$, then you can find a polynomial of degree exactly 3 that added to it will give you the zero polynomial (which is not in the set). So you might also say Axiom 5 is "sort of" satisfied.
Axiom 6 fails: for example, $x^3$ is in your set, $c=0$ is a scalar, but $0(x^3)$ is not in your set.
It's not hard to verify that Axioms 7, 8, 9, and 10 do hold.
So for 15, the axioms that fail are Axioms 1, 4, 6, and possibly 5 (depending how you interpret it).
You'll find similar problems with 16. There's a bit more to do with 17, because you also have the condition "and passes through the origin"; be sure to take that into account. Similar with 18. As for 13 and 14, I'll spill the beans and tell you that they are vector spaces: you should verify that all the axioms hold, one by one. Be sure to not verify them "by example": it's not enough to show that for particular $4\times 6$ matrices $\mathbf{u}$ and $\mathbf{v}$ you have $\mathbf{u}+\mathbf{v}=\mathbf{v}+\mathbf{u}$: you must verify it works for all possible choices of $\mathbf{u}$ and $\mathbf{v}$. If you find yourself saying "since, for example..." chances are you're doing it wrong.
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
The eight axioms define what a vector space is. If $(V,+,.)$ fails in at least one of these axioms, it's not a vector space. If $(V,+,.)$ satisfy all the axioms, it's a vector space. You can see these axioms as what defines a vector space.
a. Guess $W=\{ av+bw:a,b\in\mathbb{R}\}$ so that it's the set of combinations of $v,w\in V$ where $V$ is a vector space as I understood. $W$ is a vector space and you can prove it easly using what I wrote bellow in 3.
b. Same remark.
You can prove that $(S,+,.)$ is a vector space (i.e., satisfies all the 8 axioms) in a much easier way if you notice that $S$ is a subset of a set $V$ such as $(V,+,.)$ is a vector space. For example, we prove using the 8 axioms that $(E,+,.)$ is a vector space (there are a lot of examples like $E=\mathbb{R}^n$). Now if you notice that $V\subset E$ then $(V,+,.)$ is a vector space if and only if:
$V\neq\emptyset\,\,\,\,\,\,\,\,\,\,\,\,(1)$
$\forall u,v\in V,\forall\alpha ,\beta\in\mathbb{R},\,\alpha.u+\beta.v\in V\,\,\,\,\,\,\,\,\,\,\,\,(2)$
A good way to prove that $V\neq\emptyset$ is to prove that $0_E\in V$ where $0_E$ is the neutral element in the abelian group $(E,+)$, because if $0_E\notin V$, $V$ isn't a vector space since $(2)$ isn't correct.