One of the easiest examples I can think of for frobenius algebras is a plain ol' matrix algebra with tr : V → k as the co-unit (or equivalently, tr(a⋅b) as the frobenius form). This is enough data to generate a comultiplication δ : V → V ⊗ V. This turns out to be μ†, for multiplication μ. Is there any intuition for what this map does (aside from the obvious "do multiplication on the dual space")?
Here's how I live to think about matrices. Penrose (1971) figured out that you can draw linear algebra diagrammatically. A picture in the Penrose notation is a directed labeled graph with external leaves. The edges are labeled by vector spaces (changing the direction on an edge has the same effect as swapping the label X with the dual vector space X*), and vertices by multilinear maps. In this way, placing two edges next to each other is the tensor product. The ground field R should be drawn as an invisible edge, so that X ⊗ R = X.
So, pick your favorite finite-dimensional vector space X, and think about the types of diagrams you can draw using just it. Well, the space of matrices (what you call V) is X ⊗ X*, so it looks like two parallel lines pointed in opposite directions. Then you can check that the trace is the directed cap, the identity element (thought of as a map R → V) is the directed cup, and multiplication and comultiplication are both given by trivalent vertices.
In ASCII (ignore the weird coloring):
| | | | | | | | X = ^ , X* = v , R = [empty], V = ^ v | | | | | | | | ->- | | / \ ^ v Tr = | | I = | | ^ v \ / | | -<- | | | | | | ^ v ^ v ^ v | | | | | | mu = / _ \ delta = \ \_/ / / / \ \ \ / | | | | | | ^ v ^ v ^ v | | | | | |
Not only does the notation "explain" the comultiplication, it "proves" all the associativity and unital properties you might want. Mostly, though, I think it makes it totally clear what the Frobenius pairing (a,b) → Tr(ab) is doing. It's just the map:
->- / _ \ pair = / / \ \ | | | | ^ v ^ v | | | |
Which is just the canonical fact that (X ⊗ X*)* = X ⊗ X*. This ability to rotate X ⊗ X* is why δ = μ*.