The correspondence between edge- and vertex-labellings may work better starting from the degree 1 vertices instead of edge $n$ which would be at a random location inside the tree. For the degree 1 vertices we have a unique choice of label to assign. Remove these vertices and edges and repeat the process until the tree is a single vertex (0 edges) or a single edge. These cases correspond to enumeration of trees on labelled vertices, with the "center" of the tree being a vertex or an edge, respectively.
I'm a bit surprised that this exercise appears so early in a combinatorics book without any hints.
Regarding your first question, yes, I understand the question the same way. What I don't understand is how you originally expected us to help you in interpreting the question without linking to your book or explaining what the "first row" was. I'd suggest to always directly link to any book you're referring to if it's available online, and also not to change the spelling if you quote from a book – I searched for the problem before you provided the link and would have found it if you hadn't changed the spelling of "labeled" to "labelled" :-)
Regarding your second question: What you want to count isn't the ways of labelling the vertices of a given graph; you want to count all such graphs with all labellings for all structures; so I don't think this approach will get you very far.
The problem itself doesn't define what exactly a rooted trivalent tree is. In particular, it's not immediately clear whether the root itself can have degree $1$ or $3$ or only one of the two. Unfortunately Google Books doesn't show (at least to me) the page with Figure 14.3 that the problem refers to, which would probably resolve this. However, this page seems to indicate that the root may be required to have degree $1$, which would fit with how this book seems to be use the term, and also with this solution, which apparently makes that assumption.
The given condition is equivalent to the condition that the vertices are labelled in decreasing order as viewed from the root. Thus we're counting the number of labelled full/proper/strictly binary trees with vertex labels increasing from the root.
The recurrence
$$a_n=\frac12\sum_{k=1}^{n-1}\binom{2n-2}{2k-1}a_na_{n-k}$$
derived in the solution linked to yields the initial terms $1,1,4,34,496,\dotsc$, which lead to OEIS sequence A002105.
Best Answer
Consider the very simple tree with one vertex of degree $2$ and two vertices of degree $1$. If I don’t label the vertices, all such trees are isomorphic: for all practical purposes there is only one such tree, though I can draw it in various ways. Two of these ways are shown below:
No matter how I draw it, the vertex of degree $2$ is adjacent to each of the other two vertices.
If I label the vertices with the numbers $1,2$, and $3$, however, there are $3$ distinguishable trees: the vertex of degree $2$ can always be distinguished from the other two vertices, since they have a different degree, so labelling that vertex $1$ is different from labelling it $2$ or $3$. The three labelled trees shown below are not the same when the labelling is taken into account:
However, the first of these is indistinguishable from
which is just the same labelled tree turned around: both have vertex $2$ as the unique vertex of degree $2$, and it is adjacent to each of vertices $1$ and $3$. It’s still the same labelled tree when I draw it in either of these ways:
To give a concrete, everyday example, a family tree for one person extending back to the grandparents looks (in almost all cases) like this:
The vertices are labelled with the names of the people involved. If you shift the labels around, you get a different family tree. A is a child of B, but if you interchange the labels A and B, you have a tree showing B as the child of A, a very different thing.