See my answer here and Matthew Daly's comment. In this case, I would say (and many others) that the $ij$th entry $a_{ij}$ of $A^n$ is the number of walks of length $a_{ij}$ from $v_i$ to $v_j$, where $A$ is the adjacency matrix (More information in this stack exchange question.)
In this particular text, it looks like path is defined to be as what is otherwise referred to as a walk, and then they use simple to clarify, which is not uncommon.
I'd recommend taking a look at Douglas West's Introduction to Graph Theory if you can, specifically sections $1.1$ and $1.2$. I believe West defines a path graph in section $1.1$ and implicitly uses the "isomorphic to a path graph definition" in section $1.2$.
It's always a spanning tree.
You probably already noticed this, but for completeness: the resulting graph is acyclic, because every cycle in the original graph has been destroyed. So we need to show that the result is still connected.
Another characterization of connectivity will be useful here: a graph $(V,E)$ is connected if and only if for every nonempty $S \subsetneq V$, there is a crossing edge: an edge between a vertex in $S$ and a vertex in its complement $V \setminus S$. So let's check this for the graph after deletions.
For a given set $S$, because our starting graph was connected, there are some crossing edges. Let $e$ be the lightest of these edges. I claim that the edge $e$ is never deleted, and so there is also a crossing edge in the graph we get at the end.
For $e$ to be deleted, we'd first have to find a cycle containing it. That cycle contains at least one vertex in $S$ and at least one vertex not in $S$. Following that cycle starting from $S$, at some point we leave $S$ - but then we have to come back to $S$ by a different edge. This can happen multiple times, but even if it only happens once, we see that the cycle contains at least two crossing edges: $e$, and some other edge $e'$ (and maybe others).
Since $e$ is the lightest crossing edge, it is in particular lighter than $e'$. So it is not the heaviest edge on this cycle, and will not be deleted when we consider this cycle. The same argument holds for every cycle containing $e$, so the edge $e$ will never be deleted.
In fact, the tree $T$ we get at the end is a minimum spanning tree.
To see this, take any other spanning tree $T'$. Let $e$ be an edge of $T$ not in $T'$. Adding $e$ to $T'$ creates a cycle, and deleting any edge of that cycle would create another spanning tree. Let's add $e$ and delete the heaviest edge of that cycle.
That heaviest edge is definitely not $e$, because $e$ is not the heaviest edge of any cycle. So we added $e$ to $T'$, then deleted an edge heavier than $e$. This means that we've reduced the total weight of $T'$: therefore, $T'$ is not a minimum spanning tree. Since some minimum spanning tree must exist, it can only be $T$.
Best Answer
Node-independent, also vertex-independent, means the two paths do not share any nodes (vertices).
That is, the two paths are completely separate except for their start and end nodes. Also equivalently, the union of the paths is a circuit, and the intersection of the paths is exactly the start and end nodes.
See for example, the definitions provided at Analytic Tech's graph theory introduction or in the introduction to this paper on the connection between connectivity and vertex-independent paths.