The trace is a key concept in field theory, and ultimately this is what motivates this question.
Consider a ring $A$ such that $A \subseteq B$, and $B$ is a free $A$-module of rank $n$ (or if you rather $B$ is a finite field extension of some field $A$). Suppose $ \beta_1,...,\beta_n$ forms a basis of $B$ over $A$, then for any $b \in B$, we can write $b = \sum_{k=1}^n a_k \beta_k$ for some $a_k \in A$.
Now from here, is there a general way to construct the linear operator $M_b: B \to B$, $\big[$such that $Tr_{B/A}(b) = Tr(M_b)\big]$ from the definition of $b = \sum_{k=1}^n a_k \beta_k?$
Working through an example on paper, it would seem you require to know something about how the basis element relate through multiplication to one another. But I wondered whether there was a way around this
I think the notation $M_b$ is supposed to mean "multiplication by $b$." And indeed, the multiplication map \begin{align*} M_b: B &\to B\\ x &\mapsto bx \end{align*} is $A$-linear. (The fancy name for this is the regular representation.) More or less by definition, then $\DeclareMathOperator{\Tr}{Tr} \Tr_{B/A}(b) = \Tr(M_b)$. One can compute $\Tr(M_b)$ by finding the matrix of $M_b$ relative to some basis, but as you say, then you will need to know how to multiply basis vectors. (In general, I suppose you can write $\beta_i \beta_j = \sum_{k} c_{ijk} \beta_k$ for some $c_{ijk} \in A$ and then write everything in terms of the $c_{ijk}$.)
This requires slightly more than $B$ being a free $A$-module: we need $B$ to be an $A$-algebra, too. But this is the case when $B$ is a field (or ring) extension of a field (or ring) $A$.