Clarifying a definition of bridge I found this definition of a bridge in Graph Theory of Bondy and Murty, GTM at the page 263

Let H be a proper subgraph of a connected graph G. The set $E(G) \setminus E(H)$ can be partitioned into classes as follows.
$\vartriangleright$ For each component $F$ of $G - V(H)$, there is a class consisting of the edges of $F$  together with the edges linking $F$ to $H$
$\vartriangleright$ Each remaining edge $e$ (that is, one which has both ends in $V(H)$) defines a singleton class $\{e\}$
The subgraphs induced by these classes are the bridges of H in G

What does it means ? I don't understand how edges which has both ends in $V(H)$ are in $E(G) \setminus E(H)$ because they are supposed to be removed by hypothesis. Also why edges in F can be part of a bridge, they doesn't "link" two components ?
 A: The best way to get an intuition of bridges is to flip ahead a few pages and see how the definition facilitates its use in attempting to construct the planar embedding of a graph.  Speaking informally, let's say you wanted to try to construct a planar embedding of the Petersen graph $G$.  You might well start by choosing a large cycle $C$ of $G$ and drawing that cycle in the plane.  Now how do you divide the remaining edges into the "pieces" of $G$ that still need to be embedded?  These are the bridges.  They're similar to the components of $G-C$, but not really -- two bridges may share one or more vertices of $C$ but may be split because one is on the interior of the cycle in the embedding while the other is on the exterior.  

I don't understand how edges which has both ends in V(H) are in E(G)∖E(H) because they are supposed to be removed by hypothesis. 

You're confusing the edge sets with the vertex sets of the graphs.  To give an example that has nothing to do with bridges, let $G=K_5$, let $e$ be any edge of $G$, and let $H=G-e$.  Then $e$ is an edge that has both of its ends in $E(H)$ even though the edge itself is not in $H$.

Also why edges in F can be part of a bridge, they doesn't "link" two components ?

All edges that are not in $H$ have to be a part of a bridge.  The bridge(s) are the part(s) of $G$ that have not yet been placed in your attempted planar embedding $H$.  To go back to the example above of the Petersen graph, if you chose $C$ to be just the simple 5-cycle, then your $F$ would be the other 5-cycle plus the extra 5 edges that connected those two 5-cycles.  
