Question about a proof establishing how many lines meet $ 4 $ general lines? First of all, i'm sorry for my bad english. I'm from a foreign country.  :-)
I have some questions about a paragraph appearing in page : $111$ of the following electronic textbook : https://scholar.harvard.edu/files/joeharris/files/000-final-3264.pdf


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*The paragraph says :


There is a three dimentional family of quadratic polynomials on each $ L_i \simeq \mathbb{P}^1 $.
Each restriction map : $ H^O ( \mathcal{O}_{ \mathbb{P}^{3} } (2) ) \to H^0 ( \mathcal{O}_{L_{i}} (2) ) $ is linear, so its kernel has codimension 
$ \leq 3 $.
Since there is a 10-dimentional vector space of quadratic polynomials on $ \mathbb{P}^3 $, there is thus at least one quadric surface $ Q $ 
containing $ L_1 $ , $ l_2 $ and $ L_3 $.
By Bezout's theorem, any line meeting each of $ L_1 $ , $ L_2 $ and $ L_3 $ and thus meeting $ Q $ at least three times must be contained in $ Q $.


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*Questions :


$1)$ Why is there a family of quadratic polynomials on each $ L_i \simeq \mathbb{P}^1 $ of dimension $ 3 $ ?
$2)$ What is the interpretation of the map $ H^O ( \mathcal{O}_{ \mathbb{P}^{3} } (2) ) \to H^0 ( \mathcal{O}_{L_{i}} (2) ) $ ? what does it mean ?, and why is it linear ? Why does its kernel have codimenion $ \leq 3 $ ?
$3)$ why is the vector space of quadratic polynomials on $ \mathbb{P}^3 $ of dimension : $ 10 $ ?
$4)$ Why is, by Bezout's theorem, any line meeting each of $ L_1 $ , $ L_2 $ and $ L_3 $ and thus meeting $ Q $ at least three times, contained in $ Q $ ?.
Thanks in advance for your help.
 A: 1) is just saying that $H^0(\mathbb{P}^1,\mathcal{O}_{\mathbb{P}^1}(2))$ (global sections) is three-dimensional. Remember that the global sections of $\mathcal{O}_{\mathbb{P}^n}(d)$ are given by degree $d$ homogeneous polynomials in $n+1$ variables. In this case you can count by hand the degree 2 homogeneous polynomials in 2 variables - which has dimension 3 (you have a basis $x^2,xy,y^2$).
2) You interpret this as restricting degree 2 homogeneous polynomials on $\mathbb{P}^3$ to your copy of $\mathbb{P}^1$ embedded in $\mathbb{P}^3$. For example if $\mathbb{P}^3$ has coordinates $x_1,\ldots,x_4$ and $\mathbb{P}^1$ is embedded as the locus $x_3=x_4=0$. Then any degree $3$ homogeneous polynomial in $x_1,\ldots,x_4$ restricts to a degree $3$ homogenous polynomial in $x_1,x_2$ by substituting $x_3=x_4=0$. More generally you are considering the closed subscheme exact sequence 
$$0\rightarrow I_Y \rightarrow \mathcal{O}_{\mathbb{P}^3} \rightarrow \mathcal{O}_Y \rightarrow 0$$
where $Y$ is your copy of $\mathbb{P}^1$ lying inside $\mathbb{P}^3$, twisting by $\mathcal{O}_{\mathbb{P}^3}(2)$ and considering the corresponding map on global sections.
The kernel has codimension $\le 3$ because this is a map from a 10-dimensional vector space to a 3-dimensional vector space, so it's kernel has dimension $\ge 7$. 
3) You can count the degree 2 homogeneous polynomials with 4 variables.
4) Bezout's theorem (Corollary 2.4) tells you that if the line intersects $Q$ generically transversely they $L$ and $Q$ intersects at $2$ points counting multiplicity, because $Q$ has degree $2$ and $L$ has degree $1$. So if $L$ intersects $Q$ at $3$ points, then its intersection must be $1$-dimensional, and hence must be equal to $L$. 
