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 think you need to revise your treatment of degree when working with bilinear maps, since the notion of "bilinear" requires more information from $V \times W$ than just the fact that it is an object in the category. For a simple case, try forgetting the differentials, and just work with graded vector spaces, or comodules over $k[\mathbb{Z}]$, where $k$ is your base field. The correct notion of tensor product forces a bilinear map from $V \times W$ to respect total degree.
Best Answer
To my mind there are two classes of interesting categorical facts here, loosely speaking "additive" facts and "multiplicative" facts. Some additive facts:
Finite-dimensional vector spaces over $k$ has biproducts, and every object is a finite biproduct of copies of a single object, namely $k$. The categories with this property are precisely the categories of finite rank free modules over a semiring $R$ (the endomorphisms of the single object). These biproduct decompositions encapsulate both the idea that vector spaces have bases and that a choice of bases can be used to write linear maps as matrices.
The single object $k$ above is simple, and so every object is a finite biproduct of simple objects. The categories with this property, in addition to 1, are precisely the categories of finite rank free modules over a division semiring $R$.
Note that additive facts can't see anything about fields being commutative. The multiplicative facts can:
Finite-dimensional vector spaces over $k$ is symmetric monoidal with respect to tensor product, and is also closed monoidal and has duals (sometimes called compact closed). This observation encapsulates the yoga surrounding tensors of various types (e.g. endomorphisms $V \to V$ correspond to elements of $V \otimes V^{\ast}$), as well as the existence and basic properties (e.g. cyclic symmetry) of the trace.
The single object $k$ above is the tensor unit, and so every object is a finite biproduct of copies of the unit. Also, the monoidal structure is additive in each variable. I believe, but haven't carefully checked, that the (symmetric monoidal) categories with this property, in addition to 1 and 2 above, are precisely the (symmetric monoidal) categories of finite rank free modules over a commutative division semiring ("semifield") $R$. This encapsulates the concrete description of tensor products as a functor in terms of Kronecker products.
There's surprisingly little to say as far as fields being rings as opposed to semirings, though. This mostly becomes relevant when we reduce computing (co)equalizers to computing (co)kernels by subtracting morphisms, as in any abelian category.
Edit: I haven't mentioned the determinant yet. This mixes additive and multiplicative: abstractly the point is that we have a natural graded isomorphism
$$\wedge^{\bullet}(V \oplus W) \cong \wedge^{\bullet}(V) \otimes \wedge^{\bullet}(W)$$
where $\wedge^{\bullet}$ denotes the exterior algebra (which we need the symmetric monoidal structure, together with the existence of certain colimits, to describe). It follows that if $L_i$ are objects which have the property that $\wedge^k(L_i) = 0$ for $k \ge 2$ then
$$\wedge^n(L_1 \oplus \dots \oplus L_n) = L_1 \otimes \dots \otimes L_n.$$
Combined with the facts above this gives the existence and basic properties of the determinant, more or less. Note that the exterior algebra can be defined by a universal property, but to verify that the standard construction has this universal property we need the symmetric monoidal structure to distribute over finite colimits. Fortunately this is implied by compact closure.