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TFA specifies the bridge problem in the form where you have to "end up back home", so its conclusion is correct.

Except that, if I understand correctly, the author poses the problem such that one must 'end up back home' (that is, one must find an Eulerian circuit, not a path). If there are 2 nodes of odd degree, then you must start at one and end at the other.

I may be misremembering my graph theory class, but isn't the necessary and sufficient conditions for an eulerian path just that every vertex needs to have even degree?

I'm not sure how it could work if any nodes had odd degree.

For an Eulerian circuit, you need the graph to be connected and for every node to have even degree. For an Eulerian path (which doesn't have to start and end at the same point) you may also allow two vertices to have odd degree.

Got it, thanks!

It’s sufficient but not necessary. Consider a trivial example with three nodes arranged in a chain. The edge nodes at the end of the chain both have degree 1 but walking from one edge to another is an eulerian path.