For the concrete instances that you are considering, the finite field ${\mathbb F}_q$ and the ring of polynomials ${\mathbb F}_q[x]$, the following may help: (i) You have the elementary matrices $e_{ij}(\lambda)\in\text{GL}_n(R)$ for $\lambda$ in a ring $R$ and $i\neq j$, ranging from $1$ to $n$ and these satisfy the linearity condition $e_{ij}(\lambda)e_{ij}(\mu)=e_{ij}(\lambda+\mu)$ from where it follows that $e_{ij}(\lambda)^{-1}=e_{ij}(-\lambda)$. (ii) You also have commutator relations $[e_{ij}(\lambda),e_{k\ell}(\mu)]=\begin {cases} 1 &\text{if $j\neq k$ and $i\neq \ell$}\\ e_{i\ell}(\lambda\mu)&\text{if $j= k$ and $i\neq \ell$}\\ e_{kj}(-\mu\lambda)&\text{if $j\neq k$ and $i= \ell$}. \end{cases}$
Then, for $n\geq 3$ you define the Steinberg group $\text{St}(n,R)$ with generators abstract elements $x_{ij}(\lambda)$ subject to the linearity relations and the first two commutator relations. There is the obvious homomorphism $\varphi:\text{St}(n,R)\rightarrow \text{GL}_n(R)$ sending $x_{ij}(\lambda)$ to $e_{ij}(\lambda)$ whose image is the subgroup generated by all elementary matrices. Next pass to the direct limit over $n\rightarrow\infty$ to a homomorphism $\varphi:\text{St}(R)\rightarrow \text{GL}(R)$, whose kernel is the Milnor $K$-group $K_2(R)$. Its first property is that $K_2(R)$ is the center of the Steinberg group $\text{St}(R)$.
Next, for invertible elements of the ring you have defined the elements $g_{ij}(\lambda)$ and $h_{ij}(\lambda)$ and the first result is that if the ring $R$ is commutative and $W$ is the subgroup of the Steinberg group generated by the $g_{ij}(\lambda)$ then its image $\varphi(W)\subseteq\text{GL}_n(R)$ is the set of all monomial matrices of determinant $1$. Another relevant result is that conjugation by an element of $W$ takes each generator of the Steinberg group $\text{St}_n(R)$ to another generator. This can be made explicit and can be found with all the details in Chapter 9 of Milnor's classical "Introduction to Algebraic K-Theory" (PUP, 1971). In this same chapter Milnor introduces the Steinberg symbols $\{u,v\}=[h_{ij}(u),h_{ik}(v)]$ and shows that they are skew-symmetric and bimultiplicative. A calculation shows that if both $u$ and $1-u$ are units, then $\{u,1-u\}=1$ and for any unit $\{u,-u\}=1$. A direct consequence is that for a finite field ${\mathbb F}_q$ one has that $\{u,v\}=1$ for all $u,v\in{\mathbb F}_q^{\times}$. A little more work shows that for any field $K_2(F)$ is generated by the symbols $\{u,v\}$. For a finite field ${\mathbb F}_q$ its group of units ${\mathbb F}_q^{\times}$ is cyclic of order $q-1$ and half of the units are squares and the other half non-squares. A quick counting implies that there is a non-square $u$ such that $1-u$ is also a non-square. It follows that the group $K_2({\mathbb F}_q)$ is trivial. I hope this gets you started and then find the computation for the ring ${\mathbb F}_q[x]$, which is not as straightforward as for the field.