Disqualifier: this isn't a complete answer.
There's a basic "chalk and cheese" problem here. The "categories" that you are comparing are of two different types, although they do seem similar on the surface. On the one hand you have an honest algebraic category: that of associative algebras. But the other category (which, admittedly, is not precisely defined) is "vector spaces plus quadratic forms". This is not algebraic (over Set). There's no "free vector space with a non-degenerate quadratic form" and there'll (probably) be lots of other things that don't quite work in the way one would expect for algebraic categories. For example, as you require non-degeneracy, all morphisms have to be injective linear maps which severely limits them. You could add degenerate quadratic forms (which means, as AGR hints, that you regard exterior algebras as a sort of degenerate Clifford algebra - not a bad idea, though!) but this still doesn't get algebraicity: the problem is that the quadratic form goes out of the vector space, not into it, so isn't an "operation".
However, you may get some mileage if you work with pointed objects. I'm not sure of my terminology here, but I mean that we have a category $\mathcal{C}$ and some distinguished object $C_0$ and consider the category $(C,\eta,\epsilon)$ where $\eta : C_0 \to C$, $\epsilon : C \to C_0$ are such that $\epsilon \eta = I_{C_0}$. In Set, we take $C_0$ as a one-point set. In an algebraic category, we take $C_0$ as the free thing on one object. Then the corresponding pointed algebraic category is algebraic over the category of pointed sets (I think!).
The point (ha ha) of this is that in the category of pointed associative algebras one does have a "trace" map: $\operatorname{tr} : A \to \mathbb{R}$ given by $(a,b) \mapsto \epsilon(a \cdot b)$. Thus one should work in the category of pointed associative $\mathbb{Z}/2$-graded algebras whose trace map is graded symmetric.
In the category of pointed vector spaces, one can similarly define quadratic forms as operations. You need a binary operation $b : |V| \times |V| \to |V|$ (only these products are of pointed sets) and the identity $\eta \epsilon b = b$ to ensure that $b$ really lands up in the $\mathbb{R}$-component of $V$ (plus symmetry).
Whilst adding the pointed condition is non-trivial for algebras, it is effectively trivial for vector spaces since there's an obvious functor from vector spaces to pointed vector spaces, $V \mapsto V \oplus \mathbb{R}$ that is an equivalence of categories.
Assuming that all the $\imath$s can be crossed and all the $l$s dotted, the functor that you want is now the forgetful functor from pointed associative algebras to pointed quadratic vector spaces.
I'll denote your category of 2-vector spaces by 2Vect. By your preliminary remarks, 2Vect is actually the category of Vect-valued presheaves on Δ≤1 where Δ≤1 denotes the full subcategory of Δ on the objects [0] and [1]. Therefore, colimits in 2Vect are computed objectwise under this identification. So the functor – ⊗ X certainly has a right adjoint Hom(X, –) (by the adjoint functor theorem for locally presentable categories). This adjunction also respects the Vect enrichment.
To compute this adjoint, we can use the "Vect-enriched Yoneda lemma": writing Δi for the 2-vector space (Δi)j = HomΔ≤1([j], [i]) • ℝ, we have hom2Vect(Δi, X) = Xi as vector spaces, where hom denotes the Vect-enriched Hom. So
Hom(X, Y)0 = hom(Δ0, Hom(X, Y)) = hom(Δ0 ⊗ X, Y) = hom(X, Y)
since Δ0 happens to be the unit for this ⊗, but
Hom(X, Y)1 = hom(Δ1 ⊗ X, Y)
will have a more complicated formula which you'll have to work out (my guess is it will look similarly complicated to your expression for the other tensor product).
Best Answer
I do not think the concept of Moore-Penrose Inverse and the concept of categorical adjunction have much in common (except they both try to generalise the concept of inverse):
Equations $g = gfg$ and $fgf$ do not reaseble triangle equalities. Let me focus on the first equation. The "corresponding" triangle equality says that the composition $g \to^{\eta_g} gfg \to^{g\epsilon}g$ is the identity on $g$. Obviously $\eta_g$ is an inclusion, but generally there are no reasons for $\eta_g$ to be an isomorphism. However, in the world of 2-posets (i.e. categories enriched over posets) there is a reason. If we have a pair of 2-morphisms $\mathit{id} \le gf$ and $fg \le \mathit{id}$, then we may compose the first one on the right and the second on the left with $g$, obtaining $g \le gfg$ and $gfg \le g$, therefore $g = gfg$. Notice that I have not used any triangle equality here (just the existence of an appropriate pair of 2-morphisms).
The concept of an adjunction is inherently asymmetric --- a left adjoint is the best approximation of the identity from the left, and the right adjoint is the best approximation of the identity from the right; whilst the concept of Moore-Penrose pseudoinverse is perfectly symmetric. This means that if we had a 2-category, buit upon the category of vector spaces, where every pseudoinverse was a part of an adjunction (satisfying, perhaps, some other conditions), then every morphism would have both left and right adjoint, furthermore these adjoint functors would be isomorphic to the pseudoinverse (so isomorphic to each other). For a morphism $f$ in a 2-poset being both a left and right adjoint to $g$ is just being the inverse of $g$ --- the unit and counit from one adjunction give inequalities $\mathit{id} \le gf$ and $fg \le \mathit{id}$, whereas the unit and counit of the other adjunction give inequalities $\mathit{id} \le fg$ and $gf \le \mathit{id}$, hence $\mathit{id} = gf$ and $\mathit{id} = fg$. This fully answers your precise question, since in the category of vector spaces not every pesudoinverse is an inverse.
In any 2-category adjunctions compose --- if $f \colon A \to B \dashv f^+ \colon B \to A$ and $g \colon B \to C \dashv g^+ \colon C \to B$ then $g f \colon A \to C \dashv f^+ g^+ \colon C \to A$, but Moore-Penrose pseudoinverses --- generally --- do not. This almost answers your main question, because if $f \colon A \to B$, $g \colon B \to C$ are any maps between vector spaces then from the above property: $(gf)^+ \approx f^+ g^+$, so from this point of view pseudoinverses are not stable under isomorphisms, thus are not categorical.