From what I can dig up, given a vector bundle $E\rightarrow X$, the determinant bundle associated to this is $\Lambda^{n}E\rightarrow X$, where $n$ is the rank of $E\rightarrow X$. Is this the same thing as "the determinant of the vector bundle $E\rightarrow X$"?
If this is true, then does the statement "$E\rightarrow X$ has trivial determinant" mean that $\Lambda^{n}E\rightarrow X$ is trivializable as a vector bundle"?
The answer to both of your questions is yes! This can be seen through the construction of the top exterior bundle via transition functions.
The top exterior power of a vector bundle $\Lambda^n E$ of a rank $n$ vector bundle $E$ has transition functions $\wedge^n g_{UV}:U\cap V\to\text{GL}(\wedge^n\mathbb{C}^n) = \mathbb{C}^*$ where $g_{UV}$ are the transition functions coming from $E$. This immediately implies what we are looking at is a line bundle and the expansion is shown cleanly as; \begin{align*} (\wedge^ng_{UV})(e_1\wedge\dots\wedge e_n) &= \wedge_{i=1}^ng_{UV}(e_i) \\ &= \wedge_{i=1}^n\left(\sum_{j=1}^n g_{ji}e_j\right)\\ &= \left[\sum_{\sigma \in S_n}(-1)^{|\sigma|}\prod_{i=1}^n g_{i\sigma(i)}\right] \wedge_{i=1}^ne_i \\ &= \det(g_{UV})\cdot e_1\wedge\dots\wedge e_n. \end{align*} where the $e_i$'s denote the standard basis vectors and we have made use of the relations \begin{align*} \alpha\wedge\alpha &= 0\\ \alpha\wedge\beta &= -\beta\wedge\alpha \end{align*} for any $\alpha,\beta\in \Omega^1(E)$. This line bundle is, for now apparent reasons, known as the determinant bundle of $M$ and is also denoted $\det(E)$.
Finally, saying that $E$ having trivial determinant means that the determinant bundle of $E$ is a trivial bundle.
Note that the notion of top exterior power (for a vector space) is in fact the formal definition of determinant. That is the determinant is the unique multi-linear functional acting on $n$ vectors in an $n$-dimensional space which is alternating and whose evaluation on the standard basis is 1 (i.e. preserves the volume of the unit cube).
