For what it's worth, the construction you describe features prominently in http://front.math.ucdavis.edu/0807.4146, though that paper does not use 2-categorical language.
More generally, given a 2-category $C$ (which I usually assume is pivotal, though maybe that's not necessary here), one can construct a 1-category $D$ whose objects are 1-morphisms $f:a\to b$, and whose morphisms are "rectangles": the domain $f:a\to b$ along the bottom, the range $f':a'\to b'$ along the top, additional 1-morphisms (of $C$) $g:a\to a'$ and $h:b\to b'$ along the right and left sides, and a 2-morphism of $C$ filling in the rectangle. Composition in $D$ is given by stacking the rectangles vertically. I like to think of the pair $(g, h)$ as the (bi)grading of the morphisms of $D$. What you describe is the case where we restrict $h$ to be an identity 1-morphism (of $C$). In the paper linked to above we put an inner product on $D$ and complete it to a von Neumann algebra (in fact, a factor).
In response to David's comment below:
Modulo some details, a planar algebra is equivalent to a pivotal 2-category whose 2-morphisms are vector spaces and whose 1-morphisms are finitely generated. The standard example is constructed from a pair of factors (irreducible von Neumann algebras) $N\subset M$. From this data we construct a 2-category whose objects are $N$ and $M$, whose 1-morphisms are generated by the two bimodules $_N M_M$ and $_M M_N$, and whose 2-morphisms are intertwinors.
(So for example the 1-morphisms are
$M\otimes_N M\otimes_N\cdots\otimes_N M$, thought of as either an $N$-$N$ or $N$-$M$ or $M$-$N$ or $M$-$M$ bimodule.) You can think of the usual planar algebra definition as axiomatizing the "string diagrams" you would draw for this 2-category.
The diagrams in the paper I referred to are rotated 90 degrees from my explanation above. The left and right sides of the rectangles in the paper correspond to the $f$ and $f'$ of your (David's) original question. The tops of the rectangles corresponds to your $\phi$, and the interiors of the rectangles correspond to your $\phi^\sharp$.