In simple terms, two graphs are isomorphic to each other so long as there is a bijection between the two vertex sets and such that the bijection preserves edges. That is, two graphs $G$ and $G'$ are isomorphic if there exists a bijection, $\phi: V(G) \to V(G')$, and if that bijection also preserves edges. Explicitly, this means that if $uv$ is an edge in $E(G)$, then $\phi (u) \phi (v)$ is an edge in $E(G')$.
One can visualize a graph isomorphism in two ways.
1. Start with a graph and move around vertices in what ever way you want while keeping all the edges in tact. The result, a graph that looks different, but isn't really.
2. A re-labeling of the vertices of $G$. This produces a graph that looks the same, but the vertices are called something else.
Automorphisms are a little bit trickier. The way I look at an automorphism is a moving around of vertices (while keeping edges in tact, as in (1) above) with the caveat that the graph must look exactly the same as before. For example, take $C_4$ with vertex set $V(G) = u_1,u_2,u_3,u_4$ and edge set $E(G) = u_1u_2,u_2u_3,u_3u_4,u_4u_1$ and consider the following bijection.
1. $\phi: V(G) \to V(G')$ where $\phi$ is the permuation $(u_1u_2u_3u_4)$ that sends $u_1 \to u_2$ $u_2 \to u_3$ $u_3 \to u_4$, and $u_4 \to u_1$. (If one is unfamiliar with cycle notation). This is an isomorphism since every edge is preserved, and indeed it is also an automorphism since the resulting graph looks exactly the same as the regular graph. (the permuation above is really just a rotation by 90 degrees if you lay out the original as a "square", as is tradition.
To apply the definition in Bondy's textbook, we first have to give the vertices names so we can refer to them:
In the first graph, the external face has boundary $\{ab, ad, bc, cd\}$ and the internal face has boundary $\{ab, ad, bc, be, cd, df\}$, when regarded as edge sets. In the second graph, the external face has boundary $\{ab, ad, bc, be, cd\}$ and the internal face has boundary $\{ab, ad, bc, cd, df\}$. The drawings are not equivalent.
Note that we do not care which order the faces are written in, and which face is external. In a third drawing with both leaves on the outside, the external face would have boundary $\{ab, ad, bc, be, cd, df\}$ and the internal face would have boundary $\{ab, ad, bc, cd\}$. This would be equivalent to the first drawing.
The definition in Bondy's textbook (the edge-set definition) is not equivalent to the homotopy definition. The second example you gave illustrates that. In the edge-set definition, the two embeddings are equivalent: there is only one face, and its boundary is the set of all edges in the graph. (In fact, in the edge-set definition, every tree has a unique embedding.)
There is another combinatorial approach to embedding equivalence. You could call this the face-walk definition. For every face, you write down the closed walk that goes around its boundary in order. For example, for the first graph above, the external face has closed walk $(a,b,c,d,a)$ and the internal face has closed walk $(a,b,e,b,c,d,f,d,a)$. We say that two closed walks are equivalent if we can turn one into the other by possibly choosing a different starting point and/or reversing it, and two embeddings are equivalent if their faces have equivalent closed walks.
This definition distinguishes different embeddings of a tree, and is essentially the same as the homotopy definition. You have to be a bit careful, because as I've stated it, it's not always defined when the graph is not connected, but we can resolve that by allowing a face to have multiple closed walk boundaries.
Best Answer
The big problem with testing if two graphs are isomorphic is that there's so many potential ways to match them up. With graph drawings, there is no combinatorial explosion - at least not when the graph drawing is $2$-connected. (If not, then we can still find isomorphisms between blocks efficiently, but it might be hard to figure out which blocks correspond.)
Treat a graph drawing as a plane embedding of a graph, in which the intersections are merely another type of vertex. If the graph drawing is $2$-connected, then each face has a face walk which visits all the vertices in order around the face, which is unique up to changing the starting point and reversing the order.
Let's make two initial guesses about the drawing isomorphism:
Then if our guesses were right, the rest of the isomorphism is determined. The other face touching edge $v_i v_{i+1}$ must correspond to the other face touching edge $w_i w_{i+1}$, and knowing two corresponding vertices tells us how the rest of their face walks correspond. We can repeat this to grow the region where we know the isomorphism, one face at a time. Either we reach a contradiction (two faces that must match have different lengths, or two vertices that must match have different types), or we find an isomorphism.
We can try every possible way to make these guesses, and see if they work out. In the plane, we have a shortcut: the unbounded faces must correspond.