What does one mean when they say that a certain "subspace" of quadrics has dimension $d$?
Specifically, let $\mathbb{P}^1(\mathbb{C})\rightarrow \mathbb{P}^3(\mathbb{C})$ be given by $[X_0:X_1]\mapsto [X_0^3:X_0^2X_1: X_0X_1^2:X_1^3]=[Y_0:Y_1:Y_2:Y_3]$. The image of this map is obviously contained in the zero locus of each of the three polynomials $Y_0Y_3-Y_1Y_2$, $Y_0Y_2-Y_1^2$, $Y_1Y_3-Y_2^2$. What does it mean that the space of quadrics that contain the image is three-dimensional? Do quadrics form a vector space or something? I can understand that the set of quadrics is closed under scalar multiplication, but what about addition? Also, if there is a notion of dimension, then there must be a notion of basis. Do the quadrics defined by the polynomials above form a basis of the space mentioned? Why or why not? And what is a basis in this case?
The "space of quadrics" in $\mathbb P^3$ is the space of homogeneous polynomials in $\mathbb C[Y_0,Y_1,Y_2,Y_3]$ of degree $2$, and can be identified with the vector space of global sections $$V_2=H^0(\mathbb P^3,\mathcal O_{\mathbb P^3}(2)),$$ which has dimension $$\dim V_2=\binom{3+2}{3}=10.$$ The quadric surface corresponding to $f\in V_2$ is the projective variety $$V_+(f)=\{f=0\}\subset \mathbb P^3.$$
Let $C\subset \mathbb P^3$ be the image of the map $\mathbb P^1\to \mathbb P^3$ that you described. As you know this is a twisted cubic curve, which means it is a smooth rational degree $3$ curve in $\mathbb P^3$. Consider the exact sequence $$0\to \mathscr I_C\to \mathcal O_{\mathbb P^3}\to \mathcal O_C\to 0$$ defining the closed immersion $C\subset \mathbb P^3$. Twisting by $\mathcal O_{\mathbb P^3}(2)$ and taking global sections yields a short exact sequence of vector spaces $$0\to H^0(\mathbb P^3,\mathscr I_C(2))\to V_2\to H^0(\mathbb P^3,\mathcal O_C(2))\to 0.$$ The kernel, $H^0(\mathbb P^3,\mathscr I_C(2))$, is the space of quadrics containing $C$. Indeed, the surjection $V_2\to H^0(\mathbb P^3,\mathcal O_C(2))$ is the restriction map $f\mapsto f|_C$, and a surface $V_+(f)$ contains $C$ if and only if its defining equation vanishes identically on $C$, i.e. $f|_C=0$. Since $$H^0(\mathbb P^3,\mathcal O_C(2))\cong H^0(\mathbb P^1,\mathcal O_{\mathbb P^1}(6))\cong\mathbb C^7,$$ the kernel $H^0(\mathbb P^3,\mathscr I_C(2))$ is indeed $3$-dimensional, generated by the sections you mention.
Added (Thanks Georges!): To show $V_2\to H^0(\mathbb P^3,\mathcal O_C(2))$ is surjective, we have to show that for every homogeneous polynomial $h(u,v)$ of degree $6$, there is a homogeneous polynomial $g(Y_0,Y_1,Y_2,Y_3)$ of degree $2$ such that $h(u,v)=g(u^3,u^2v,uv^2,v^3)$. The space containing the $h$'s is spanned by the monomials $h_i(u,v)=u^iv^{6-i}$, for $0\leq 0\leq 6$. We can divide by $3$ and write $$i=3p+q,\qquad 6-i=3p'+q',$$ with $0\leq q,q'\leq 2$ and $q+q'\equiv 0\bmod 3$. Then $h_i(u,v)=(u^3)^p(v^3)^{p'}u^qv^{q'}$. This direct argument can be found in this paper by Ottaviani, but generalizes to the statement that the rational normal curve in $\mathbb P^n$ is projectively normal, that is, it is normal and satisfies $H^1(\mathbb P^n,\mathscr I(d))=0$ for all $d\geq 0$. Another class of projective varieties $X\subset \mathbb P^r$ for which the restriction maps $H^0(\mathbb P^r,\mathcal O_{\mathbb P^r}(d))\to H^0(X,\mathcal O_{X}(d))$ are onto is that of complete intersections (which the twisted cubic is not).