How to see if a node is "part of" a larger way

Yes, please use way(bn).

A simple example. We have queried for all the barriers in Watford near London:

area[name="Watford"];
node[barrier](area)(51.5,-1,52,0);
out;

The query finds 84 or so.

Now we replace the output in line 3 by a couple of extra lines:

area[name="Watford"];
node[barrier](area)(51.5,-1,52,0);
way(bn)->.wy;
node._(w.wy);
out;

In detail: way(bn)->.wy queries for all the ways that have the just found nodes as members node._(w.wy) is composed of the two conditions that the node must be from the previous result (in this case line 2, since line 3 writes to elsewhere) and that it must be a node referenced by a just found way (from set wy, filled in line 3)

Thus we see only those nodes that are member of a way.

Further conditions on the interesting way can be injected in line 3:

area[name="Watford"];
node[barrier](area)(51.5,-1,52,0);
way(bn)[highway=cycleway]->.wy;
node._(w.wy);
out;

Now the ways that our nodes of interest must be members of are restricted to those with the tag highway=cycleway.

It gets more tricky for the related question: What are the nodes of a way that are also connected to other ways? As an example, the query for the street named Hyde Vale ...

( way[name="Hyde Vale"];
  node(w); );
out;

... becomes:

( way[name="Hyde Vale"]->.ways_to_exclude;
  node(w.ways_to_exclude)->.nodes_of_interest; );
( way(bn.nodes_of_interest)->.all_wy; - .ways_to_exclude; );
node.nodes_of_interest(w);
out geom;

We divert this time already the ways of initial interest through a named set ways_to_exclude because we need them again to remove them from the list of eligible ways. Other than that, lines 3 and 4 do the exercse to walk from the nodes to the ways and back to their nodes; and line 4 ensures that we only consider nodes that we had found already. But to avoid that all nodes qualify by being part of the street Hyde Vale (and potentially no other way) we explicitly remove these ways with the difference statement in line 3.

Take a look at the recurse filters. There's also an operator, <, that will find all parents of the input set.

The recurse filter:

https://wiki.openstreetmap.org/wiki/Overpass_API/Overpass_QL#Recurse_.28n.2C_w.2C_r.2C_bn.2C_bw.2C_br.29

Simple example: https://overpass-turbo.eu/s/NRJ