Number of Labels used in reduction of Isomorphism of Labelled Graph to Graph Isomorphism From "Lecture Notes in Computer Science" by Christoph M. Hoffmann ,

Assume that both $X$ and $X'$ have $n$ vertices. We plan to code the
  graph labels as suitable subgraphs which we attach to the vertices of
  $X$ and of $X'$. In time polynomial in the length of the input we can rename 
  the labels and may assume, therefore, that $L = \{1 ..... k\}$ is
  the set of labels assigned by $\lambda$ and $\mu$. Note that $k\leq2n$ .We describe
  how to construct $Z$ from $(X,\lambda)$. The construction of
  $Z'$ is done in the same way. Let $X = (V,E), V = \{v_1..... v_n\}$.
  Intuitively, we obtain $Z$ from $X$ by attaching to each vertex $v_i$ a
  complete graph with $n + r$ vertices, where $\lambda (v_i) = r$. Note that $r\leq2n$.
  The vertices of the attached complete graph will be, for $v_i$, $\{v_{(i,1)},..... v_{(i,n+r)} \}$
   The subgraph is atached to $v_i$ by an edge $(v_i, v_{i,1})$.
It is easy to see that $(X,\lambda)$ is
  isomorphic to $(X,\mu)$ iff $Z$ is isomorphic to $Z'$. Since there are at most
$2n$ distinct labels, the graphs $Z$ and $Z'$ have no more than
  $2n^2+n$ vortices each and can thus be constructed in
  polynomial time.

the definition of graph label is given as-

Let $X = (V,E)$ be a graph, $\lambda$ is  a mapping from $V$ onto a set $L = \{ l_1,...l_k\}$.
   Then the pair  $(X,\lambda)$ is a labelled graph.

Question 1: Why $k\leq 2n$ which implies $r\leq2n$?   It seems that it should be $k\leq n$ since the passage of wikipedia here tells

When used without qualification, the term labeled graph generally
  refers to a vertex-labeled graph with all labels distinct. Such a
  graph may equivalently be labeled by the consecutive integers $\{1, …,
> |E |\}$, where $|E |$ is the number of vertices in the graph.

Question 2:  What is the explanation of -

Since there are at most
  $2n$ distinct labels, the graphs $Z$ and $Z'$ have no more than
  $2n^2+n$ vortices each and can thus be constructed in

Thanks in advance.
 A: My interpretation:
The Confusion arises, here-

Assume that both $X$ and $X'$ have $n$ vertices. We plan to code the
graph labels as suitable subgraphs which we attach to the vertices of
$X$ and of $X'$. In time polynomial in the length of the input we can rename
the labels and may assume, therefore, that $L = \{1 ..... k\}$ is
the set of labels assigned by $\lambda$ and $\mu$. Note that $k\leq2n$ .

One needs to note that-

we can rename the labels and may assume, therefore, that $L = \{1 ..... k\}$ is the set of labels
assigned by $\lambda$ and $\mu$.

So, if you have $k\leq n$, then there is a guarantee or certainty,that there exists an Isomorphism  from $X$ to $X'$ already, since, if $k\leq n$ then it is certain that $\exists i,j$ such that $ \lambda(v_i)=r=\mu(v_j)$ where $ 1\leq i,j \leq n$.
But if $k\leq 2n$, then $\lambda(v_i)$ may take $n$ different label, and  $\mu(v_j)$ may take $n$ different label.
For example, consider Cyclic Graph $C_4$ and Complete Graph $K_4$. Add 1 vertex  to 1st,2 vertices  to 2nd,3 vertices  to 3rd, 4 vertices  to 4th vertex   of $C_4$ and $K_4$ and rename them as $C'_4$ and $K'_4$. Now you have to use 14 different labels in $C'_4$, and none of them can be used in  $K'_4$. Note that the number of vertices is same in both $C'_4$ and $K'_4$. Here, $|L|=2n=28$;
So, $k\leq2n$.
A: I think the answer of Jim is good. $k\leq 2 n$ because you have to consider both $\mu$ and $\lambda$. If they have different labels the total number of unique label can be $2n$. (Of course one can argue that if $\lambda$ and $\mu$ does not share the same set of labels that $(X,\lambda)$ and $(X',\mu)$ have no chance to be isomorphic, thus you could first look whether $\mu$ and $\lambda$ share the same labels and then consider $k\leq n$. But that is not the point of the proof here.)
For your second question it is an upper bound of the worst case scenario which gives the bound $2n^2+n$.
The intuition of the proof is to encode the labels of $(X,\lambda)$ and $(X',\mu)$ by connecting the vertices to new fresh vertices.
To do this they assume that the labels are between $1$ and $2n$ (which is the first part that confused you). Then they encode that a vertex $v$ has label $j$ by connecting $v$ in $Z$ to $j$ new vertices (note that only $v$ will be connected to this vertices).
So you have to introduce $\lambda(v)$ new vertices for each $v\in X$ in order to build $X$. Thus in $X$ you have exactly $n+\sum_{v\in X}\lambda(v)$ vertices. but you know that $\lambda(v)\leq 2n$ thus you get:
\begin{align}
|Z| &=n+\sum_{v\in X}\lambda(v) \\
& \leq n+\sum_{v\in X}2n = n+n2n = 2n^2+n  
\end{align}
