Completing the solution, lipschitz maps inducing other maps Let $(X, d)$ be a metric space, $(E, || \cdot ||)$ a Banach space, $(AE(X), || \cdot ||)$ - as described below.
I've already proven that for any Lipschitz function $u: X \rightarrow E $ there exists a unique linear, continuous map $AE(u) : AE(X) \rightarrow E$ such that $AE(u) (m_{xy}) = u(x)-u(y), \ x,y \in X$, although I have problems proving linearity.
But the main trouble is deducing from the above that any Lipschitz function $v : X \rightarrow Y$ between two nonempty metric spaces  induces a linear, continuous function $AE(v) : AE(X) \rightarrow AE(Y)$.
Could you help me with that?
$AE_0(X) = \{ u : X \rightarrow \mathbb{R} \ : \ u^{-1} (\mathbb{R} \setminus \{0 \}) \ \  \text{is finite}, \ \sum_{x \in X} u(x)=0  \}$, 
for $x,y,z \in X, \ x \neq y, \ m_{xy} \in AE_0(X), \ \ m_{xy} (x)=1, \ m_{xy}(y)=-1, \ m_{xy}(z)=0$ for $z \neq x, y$ and $m_{xx} \equiv 0$
for $u \in AE_0(X), \ \\ ||u||_d = \inf _{n \ge 1} \{ \sum_{k=1}^n |a_k| d(x_k, y_k) \ : \ u= \sum_{k=1}^n a_km_{x_k, y_k}, a_k \in \mathbb{R}, x_k, y_k \in X\}$
We denote the completion of $AE_0 (X)  \ \text{by} \ AE(X)$
 A: The map $AE(v)$
Let's begin by reviewing the meaning of notation. The space $AE_0(X)$ consists of formal linear combinations $\sum c_k x_k$ with real coefficients summing to zero (equivalently, signed atomic measures). If we think of positive $c_k$ as production quantity and negative $c_k$ as consumption, then the norm on $AE_0$ is the infimal transportation cost  of delivering the product from producers to consumers. 
A Lipschitz map $v:X\to Y$ pushes forward the elements of $AE_0(X)$, like any measure gets pushed under a map. Specifically, the image of $\sum c_k x_k$ is $AE(v)(u)=\sum c_k v(x_k)$, possibly after combining the coefficients if some $x_k$ are mapped to the same point.  Any transportation plan $u=\sum_{k=1}^n a_km_{x_k, y_k}$ gets pushed to $AE(v)(u)=\sum_{k=1}^n a_k m_{v(x_k), v(y_k)}$ with the cost at most $L$ times the original cost, where $L$ is the Lipschitz constant of $v$. The factor of $L$ comes from $d(v(x_k), v(y_k))\le Ld(x_k,y_k)$.
So, you have a continuous linear map from a dense subspace of $AE(X)$ into $AE(Y)$. Such a map has a unique extension to $AE(X)$. 
Linearity of $AE(u)$
I don't see why you would need $AE(u)$ for the above problem; I did not use it. And the notation $AE(u)$ is poorly chosen, because it clashes with $AE(v)$ introduced below (a Banach space is also a metric space). 
Anyway, the map should first be defined on $AE_0(X)$ as 
$$\sum c_k x_k \mapsto \sum c_k u(x_k)$$
where the sum on the right is no longer a formal linear combination, but an actual sum in $E$. The linearity should be clear (in this notation), and the continuity follows from the Lipschitz property of $u$:
$$\left\|\sum c_k u(x_k)\right\|_E\le L \|\sum c_k x_k\|_d $$
Hence, there is a unique continuous extension to $AE(X)$.
