Contrary to what Lengyel claims, Dorst et al.'s defintions of scalar products and contractions are mathematically justified: they arise from the exclusive use of the twisting symmetry on the category of graded modules, as done for example in Chapter 4 and onward of Helmsletter and Micali's book Quadratic Mappings and Clifford Algebras. By contrast, Lengyel's definitions arise from an additional use of the ordinary symmetry on the category of graded modules.
Moreover, there is no error in Dorst et al.'s (or Lounesto's) assertion that $a\rfloor(B\wedge C)=a\rfloor B+(-1)^{\deg B}B\wedge(a\rfloor C)$. Lengyel contradicts himself when claiming there is such a mistake: he claims that $(B\wedge C)\lfloor a=B\lfloor a+(-1)^{\deg B}\wedge(C\lfloor a)$ is correct in his notation, but the right contraction $B\lfloor A$ in his notation is the same as the standard left contraction $\tilde A\rfloor B$ with the reverse $\tilde A$ of $A$. In particular, the two identities are the same (and in fact both correct).
Finally, there is a mathematical basis for having contractions given by components of the geometric product, namely the the theory of deformations of Clifford algebras, which is worked out in full detail in Helmsletter and Micali's book. For Lengyel to claim otherwise is to ignore the larger algebraic context of geometric algebras.
Lengyel does have a point in that there is another inner product on the exterior algebra in addition to the scalar product of the geometric algebra. Consequently, taking the scalar product, or the other inner product, with a pseudoscalar results in two different dualities on the geometric algebra. Only the latter seems to have good geometric properties, matching the Hodge dual of exterior algebra.
I am inclined to agree with Lengyel that the absence of this inner product and corresponding correct expression of Hodge duality is a gap in the foundations of geometric algebra limting its proper geometric use. However, the existing foundations concerning the scalar product and contractions are not poor, but part of the larger algebraic context of Clifford algerbas.
Regarding Dorst et al.'s claims on transform composition and projective geometric algebra, Lengyel may be correct but I do not know to evaluate those claims.
In more detail on how Lengyel's definitions arise from using ordinary instead of twisting symmetry, recall that the twisting symmetry $A\otimes B\to B\otimes A$ on graded modules is given by mapping homogeneous elements according to $a\otimes b\mapsto(-1)^{\deg a\deg b}b\otimes a$, whereas the ordinary symmetry is given by $a\otimes b\mapsto b\otimes a$. Their composite is the twisting action on tensor products given by $a\otimes b\mapsto(-1)^{\deg a\deg b}a\otimes b$. Consequently, any graded algebra structure given by $A\otimes A\to A$ (or graded coalgebra structure given by $A\to A\otimes A$) determines three additonal graded (co)algebra structures: the opposite, twisted, and twisted opposite obtained by composing with the symmetry, twist, or twisted symmetry of tensor products.
Now, any bilinear form on a module $M$, when identified with a linear map $b\colon M\to M^*$ to its dual $M^*$, induces a bilinear form on its exterior algebra $\Lambda M$ that vanishes on elements of different degrees, and corresponds to the unique algebra homomorphism $\Lambda M\to(\Lambda M)^*$ extending the linear composite $M\to M^*\to \Lambda^* M$ to the graded algebra $\Lambda^* M$ of alternating maps on $M$. Such an extension exists (and is unique) because the algebra structure on $\Lambda ^*M$ guarantees $(m^*)^2=0$ for any image $m^*\in(\Lambda M)^*$ of $m\in M$ under that linear composite.
Explicitly, the algebra structure on $(\Lambda M)^*$ is given by $(f\wedge g)(m_1\wedge\cdots\wedge m_n)=\sum_s\DeclareMathOperator\sgn{sgn}\sgn(s)f(m_{s(1)},\dots, m_{s(i)})g(m_{s(i+1)}),\dots,g(m_{s(n)})$ where $f$ is an $i$-form, $g$ an $n-i$-form, $m_i\in M$, and $s$ ranges over permutations satisfying $s(1)<\cdots<s(i)$ and $s(i+1)<\cdots<s(n)$, and $\sgn(s)$ is the sign of the permutation. It follows that the induced bilinear form on $\Lambda M$ is given by $b(m_1\wedge,\cdots,\wedge m_n,m'_1\wedge\cdots m'_n)=\det(b(m_i,m'_j))$. Applied to an inner product on $M$, this induces Lengyel's preferred inner product on $\Lambda M$.
But $(m^*)^2=0$ in $\Lambda^*M$ implies $(m^*)^2=0$ also for the opposite, twisted, and twisted opposite algebra structures on $\Lambda^*M$. Thus, there are a priori another three induced bilinear forms on $\Lambda M$. Fortunately, $\Lambda^*M$ is twisted commutative, i.e. is the same as its twisted opposite, i.e. its opposite is the same as its twist. So there is one other bilinear form induced by the opposite algebra $\Lambda^\wedge M$ given by $b^\wedge(m_1\wedge\cdots\wedge m_n,m'_1\wedge\cdots\wedge m'_n)=\det(b(m_{n-i},m'_j))=(-1)^{n\choose2}B(m_1\wedge\cdots\wedge m_n,m'_1\wedge\cdots\wedge m'_n)$. Applied to an inner product on $M$, the inner product induced in this way is the scalar product of the correponding Clifford algebra.
The twisting symmetry first arises because of the fundamental result that the Clifford algebra $\DeclareMathOperator\Cl{Cl}\Cl((M,q)\perp(M',q'))$ of an orthogonal sum $(M,q)\perp(M',q')=(M\oplus M',q+q')$ of quadratic modules is the twisted tensor product $\Cl(M,q)\hat\otimes\Cl(M',q')$ of the Clifford algebras of those quadratic modules.
The diagonal map $M\to M\oplus M$ underlies morhpisms of quadratic modules $(M,q)\to(M,0)\perp(M,q)$ and $(M,q)\to(M,q)\perp(M,0)$, making $(M,0)$ into a cocommutative coalgebra over which $(M,q)$ has compatible left and right comodule structures. Consequently, the exterior algebra $\Lambda M=\Cl(M,0)$ is a cocommutative coalgebra in the category of graded algebras with twisted tensor product over which each Clifford $\Cl(M,q)$ algebra has compatible left and right comodule structures.
Now on the one hand, the dual $A^*$ of a graded module $A$ admits two morphisms $A^*\otimes A^*\to(A\otimes A)^*$ that are twists of one another. Each arises from following the ordinary or twisted symmetry $A^*\otimes A^*\otimes A\otimes A\to A^*\otimes A\otimes A^*\otimes A$ with the product of evaluation maps from $A^*\otimes A$. Call the former the ordinary morphism $A^*\otimes A^*\to(A\otimes A)^*$ and the latter the twisted morphism.
On the other hand, when $A\to A\otimes A$ is a coalgebra structure on $A$, the composition of $A^*\otimes A^*\to(A\otimes A)^*\to A^*$ is an algebra structure on $A^*$. Moreover, the opposite, twisted, and twisted oposite coalgebra structures on $A$ lead to the opposite, twisted, and twisted opposite algebra structures on $A^*$. Thus if $A\to A\otimes A$ is twisted cocommutative, then $A^*\otimes A^*\to A^*$ is also twisted commutative, so that its opposite and twisted structures are the same.
It follows that the twisted algebra structure on $\Lambda^*M$ arises from using the coalgebra structure on $\Lambda M$ and the twisted morphism $(\Lambda M)^*\otimes(\Lambda M)^*\to(\Lambda M\otimes\Lambda M)^*$.
Since $\Cl(M,q)$ is compatibly a left and right comodule over $\Lambda M$, it is also a bimodule over the algebra $\Lambda^*M$. Composing with the algebra homomorphisms $\Lambda M\to\Lambda^*M$, one obtains two structures of $\Cl(M,q)$ as a $\Lambda M$-bimodule, depending on whether $\Lambda^*M$ was given the ordinary or twisted algebra structure. The one arising from the twisted algebra structure consists of the left and right contractions defined by Dorst et al. The one arising from the ordinary one consists of Lengyel's left and right contractions.
Finally, since the twisted algebra structure is simply the opposite one, the two $\Lambda M$-bimodule structures on $\Cl(M,q)$ are canonical related by being opposite to one another, which explains why Lengyel's right contraction is given by Dorst et al.'s left contraction with a reversed element.
