Extension of a linear map For any $x \in X$, define the set $\mathcal{F}(X) = \overline{span \{ \delta_x : x \in X \}}$ where $\delta_x(f)=f(x)$ for all $f \in$ $Lip_0(X)$. 
The set Lip$_0(X)$ is the set of all real-valued Lipschitz functions which vanish at $0$.
Note that $\delta_x$ is an evaluation functional on Lip$_0(X)$.

Suppose $X$ and $Y$ are Banach spaces. Let $L:X \rightarrow Y$ be a
  Lipschitz map such that $L(0)=0$. Then there exists a unique linear
  map $\hat{L}: \mathcal{F}(X) \rightarrow \mathcal{F}(Y)$ such that
  $\hat{L}\delta_X = \delta_YL$

The statement above is Lemma $2.2$. 
The following is its proof:
The linear map $L^\# : Lip_0(Y) \rightarrow Lip_0(X)$ defined by $L^\#(F) = F \circ L$ is pointwise-to-pointwise continuous, hence there is a linear map $\hat{L}$ between the preduals such that $\hat{L^*} = L^\#$. It is clear that $\| L^\# \| = Lip(L)$, and $\| \hat{L} \| = \| \hat{L^*} \| = \| L^\# \|$. The other assertions are clear.
Question:
$(1)$ To show the map $L^\#$ is pointwise-to-pointwise continuous, do I show that if $\lim_{n \rightarrow \infty}{F_n}=F$, then $ \lim_{n \rightarrow \infty}{L^\#F_n}= L^\#F$?
$(2)$ Why there exists a linear map $\hat{L}$ after showing that $L^\#$ is point-to-pointwise continuous?
$(3)$ Is that a typo in $\| L^\# \| = Lip(L)$? I think it should be $\| L^\# \| = \| L \|_{Lip}$. After that, how to show that $\| L^\# \| = \| L \|_{Lip}$?
$(4)$ How to show that the map $\hat{L}$ is unique?
 A: The following does not answer your questions, it rather is an alternative (clearer?) proof to lemma 2.2, without any need for preduals.
First, note that $Lip_0 (X)$ is a Banach space with the norm given by $\| f \| = Lip(f)$, where $Lip(f) = \inf \{ M > 0 \mid |f(x) - f(y)| \le M \| x - y \| \ \forall x, y \in X \}$ (the Lipschitz constant of $f$). This means that in this topology $f_n \to f \iff Lip(f_n - f) \to 0$.
Note that if $f_n, f \in Lip_0 (X)$ with $f_n \to f$ (in the topology given by the above norm), and $x \in X$, then $| (f_n - f) (x) | = | (f_n - f) (x) - (f_n - f) (0) | \le Lip(f_n - f) \| x - 0 \| = Lip(f_n - f) \| x \| \to 0$, which can be rewritten as $\delta _x (f_n) \to \delta _x (f)$, so $\delta _x \in Lip_0 (X) ^*$, or equivalently $\mathcal F (X) \subset Lip_0 (X) ^*$.
Define $\tilde L : Lip_0 (X) ^* \to Lip_0 (Y) ^*$ by $\tilde L (\omega) (f) = \omega ( f \circ L )$ (where $\omega \in Lip_0 (X) ^*$ and $f \in Lip_0 (Y)$). Linearity is clear. Let us show continuity: if $\omega _n \to \omega$ in norm, then $\omega _n (g) \to \omega (g) \ \forall g$; taking $g = f \circ L$ closes the argument. Let $\hat L = \tilde L \big| _{\mathcal F (X)}$. It remains to show that $\text{range} \hat L \subset \mathcal F (Y)$.
Note that $\hat L (\delta _x) (f) = \delta _x (f \circ L) = f(L(x)) = \delta _{L(x)} (f)$ or equivalently $\hat L (\delta _x) = \delta _{L(x)} \in \mathcal F (Y)$, which readily implies that $\hat L (\mathcal F (X)) \subseteq \mathcal F (Y)$ by using linear combinations of evaluation functionals and then taking limits of sequences of such combinations.
Applying $\hat L \circ \delta _X = \delta _Y \circ L$ in an arbitrary point $x \in X$ gives $\hat L (\delta _x) = \delta _{L(x)}$ which has been seen to be true in the paragraph above, so the required commutativity property is true. To show uniqueness, assume that there exist $K \ne \hat L$ with the same properties as $\hat L$. In particular, since $K \circ \delta _X = \delta _Y \circ L$, you would get $K (\delta _x) = \delta _{L(x)} = \hat L (\delta _x)$, so (by linearity) $K - \hat L = 0$ on the linear (but not topological yet!) span of $\{ \delta _x \} _{x \in X}$. Since $\hat L$ has been proven continuous and $K$ is assumed continuous, then $K - \hat L$ is a continuous linear operator, so it can be uniquely extended from the linear span of $\{ \delta _x \} _{x \in X}$ to $\mathcal F (X)$ by the usual density-related argument, and since it extends $0$, this extension will also be $0$, so $K = \hat L$.
