Scalar product of vectors over polynomial rings
To encode vectors following that technique, you just have to generalize it.

For any two $(d+1)$-dimensional vectors $a = (a_0, a_1, ..., a_d)$ and $b = (b_0, b_1, ..., b_d)$, define the polynomials $a(x) = a_0 + a_1x + ... + a_dx^d$ and $b(x) = b_d + b_{d-1}x + b_{d-2}x^2 + ... + b_0x^d$. 

Then, when multiplying $a(x)$ by $b(x)$, among all the coefficients that we multiply, we will multiply $a_0$ by $b_0x^d$, also $a_1x$ by $b_1x^{d-1}$, $a_2x^2$ by $b_2x^{d-2}$, and so on... Notice that all that multiplications are of the form $a_kx^k \cdot b_kx^{d-k}$, which result in $a_kb_kx^d$. Therefore, if we look to the $d$-th coefficient of the product polynomial we will see 

$$\left( \sum_{k=0}^d a_kb_k \right) x^d$$

and that summation is exactly the inner product between the vectors $a$ and $b$.

It is worth noting that if this summation is greater than the module of the plaintext ring (usually a parameter $t$ or $p$), than the decoding will not work. Also, that representation is only possible if that value $d$ is smaller than the degree of the polynomial used to define the quotient ring (usually a cyclotomic polynomial).