Let $F_{27}=\{0, \alpha, \dots, \alpha^{26}=1\}$ and $B=\{1, \alpha, \alpha^2\}$ be a basis of $F_{27}$ over $F_3$ then an element $\alpha^k = c_1+c_2\alpha+c_3\alpha^2$ where $1\le k\le 26$.
To determine $c_1,c_2$ and $c_3$ we know $\exists$ some $\beta \in F_{27}$ such that $Tr(\beta\alpha^k) = c_i$ where $1\le i\le 3$. My confusion is which $\beta$ should be selected because for different $\beta$ the co-efficient will be $\{0,1,2\}$.
You cannot just choose a $\beta$. In fact, there are three elements $\{\beta_0,\beta_1,\beta_2\}$ that constitute the dual basis of $B = \{\alpha^0,\alpha^1,\alpha^2\}$. The dual basis has the property that for $m,n \in \{0,1,2\}$ $$\operatorname{Tr}(\beta_m\alpha^n) = \begin{cases}1, & m=n,\\0, & m \neq n, \end{cases} \qquad m,n \in \{0,1,2\}$$ Since the trace is a linear function (over $\mathbb F_3$) from $\mathbb F_{27}$ to $\mathbb F_3$, we have that $$\operatorname{Tr}(\beta_m\alpha^k) = \operatorname{Tr}(\beta_m(c_0+c_1\alpha+c_2\alpha^2))= \sum_{n=0}^2 c_n\cdot\operatorname{Tr}(\beta_m\alpha^n) = c_m, \quad m\in \{0,1,2\}.$$ In other words, three different "$\beta$'s are needed to get $c_0, c_1, c_2,$ not just a single $\beta$ as the last sentence in your question seems to hint at. Also, the same dual basis is used regardless of the value of $k$.