The definition of the dot product of a second order tensor with a vector is: $$ (\mathbf{a} \otimes \mathbf{b}) \cdotp \mathbf{x} = (\mathbf{b} \cdotp \mathbf{x})\mathbf{a} $$ The definition of the dot product of a third order tensor with a vector is: $$ (\mathbf{a} \otimes \mathbf{b} \otimes \mathbf{c}) \cdotp \mathbf{x} = (\mathbf{c} \cdotp \mathbf{x}) (\mathbf{a} \otimes \mathbf{b}) $$ Is it the case that: $$ \mathbf{x}^T \cdotp (\mathbf{a} \otimes \mathbf{b}) =(\mathbf{x}^T \cdotp \mathbf{a}^T)\mathbf{b}^T$$ I believe the transpose on $\mathbf{a}$ is needed because the dot product must be either between two vectors or two co-vectors. I also believe the transpose on $\mathbf{b}$ is needed to make the shape of the resulting vector come out right. Like in matrix multiplications where a row vector times a matrix gives a row vector. But I'm not sure whether $\mathbf{x}^T \cdotp (\mathbf{a} \otimes \mathbf{b})$ even makes sense conceptually or if only makes sense to write $\mathbf{x}^T \cdotp (\mathbf{a}^T \otimes \mathbf{b}^T)$?
Is it the case that:
$$ \mathbf{x}^T \cdotp (\mathbf{a} \otimes \mathbf{b} \otimes \mathbf{c}) = (\mathbf{c} \cdotp \mathbf{x}) (\mathbf{a} \otimes \mathbf{b})^T = (\mathbf{c} \cdotp \mathbf{x}) (\mathbf{a}^T \otimes \mathbf{b}^T) $$
This is a long comment rather than an answer.
Take your first definition, $$ (\pmb{a} \otimes \pmb{b}) \cdot \pmb{x}= (\pmb{b} \cdot \pmb{x}) \pmb{a} $$ This only makes sense if, somehow, $$ (\pmb{a} \otimes \pmb{b}) \cdot \pmb{x} = (a^i \pmb{e}_i \otimes b_j \pmb{e^j}) x^k \pmb{x}_k = a^i \pmb{e}_i (b_jx^j) $$ i.e., as if by magic, $\pmb{b}$ is a covector (in Euclidean space you could call this $\pmb{b}^T$ but then you would simply write $\pmb{a}\pmb{b}^T\pmb{x}$ as @Kurt G. suggested). Then if we write $(\pmb{a} \otimes \pmb{b}) \cdot \pmb{x}^T$ using the same definitions, $$ (a^i \pmb{e}_i \otimes b_j \pmb{e^j}) x_k \pmb{x}^k = (a^ix_i) \otimes b_j \pmb{e^j} $$ or $(\pmb{x}^T\pmb{a}) \pmb{b}^T$.
Your second definition is even more difficult to decipher. You write, $$ (\pmb{a} \otimes \pmb{b} \otimes \pmb{c}) \cdot \pmb{x}= (\pmb{c} \cdot \pmb{x}) \pmb{a} \otimes \pmb{b} $$ so $\pmb{c}$ is also a covector and we have, $$ (\pmb{a} \otimes \pmb{b} \otimes \pmb{c}) \cdot \pmb{x} = (a^i \pmb{e}_i \otimes b_j \pmb{e^j} \otimes c_k \pmb{e^k}) x^l \pmb{e}_l = (c_k x^k) (a^i \pmb{e}_i \otimes b_j \pmb{e^j}) $$ Then, $$ (\pmb{a} \otimes \pmb{b} \otimes \pmb{c}) \cdot \pmb{x}^T = (a^i \pmb{e}_i \otimes b_j \pmb{e^j} \otimes c_k \pmb{e^k}) x_l \pmb{e}^l = (c_k a^k) (b_j \pmb{e^j} \otimes c_k \pmb{e^k}) $$ i.e. using your notation, $$ (\pmb{a} \otimes \pmb{b} \otimes \pmb{c}) \cdot \pmb{x}^T = (\pmb{a} \cdot \pmb{x}) (\pmb{b} \otimes \pmb{c}) $$