Pretty much everything in the title, I feel like this is a stupid question though I'm struggling. I don't understand why K3 surfaces can be given with the two equivalent definition where one states that $K_X$ is trivial while the other states that there exists a nowhere vanishing holomorphic 2-form. I've tried looking at the definition of the canonical bundle, but the fact that it is trivial (i.e all of its sections are the zero space) leads me nowhere.
Thank you very much indeed I hope I've explained my doubt thoroughly
You are confused about what it means for the canonical bundle $K_X$ to be trivial. It does not mean that $H^0(X,K_X)=\{0\}$. It means that $K_X\cong\mathcal{O}_X$. In particular, when $\dim_\mathbb{C}X=2$, we have $K_X=\wedge^2T^*X^{1,0}$, where $T^*X\otimes\mathbb{C}=T^*X^{1,0}\oplus T^*X^{0,1}$ is the decomposition of the complexified tangent bundle into its holomorphic and anti-holomorphic parts. In particular, $T^*X^{1,0}$ can be given the structure of a holomorphic vector bundle over the complex manifold $X$, and so can $\wedge^2T^*X^{1,0}$. Its sections (viewed as a holomorphic vector bundle) are holomorphic $2$-forms. The existence of a holomorphic trivialisation $K_X\cong\mathcal{O}_X$ implies the existence of a nowhere vanishing section of $K_X=\wedge^2T^*X^{1,0}$, in the same way that an orientation of a manifold $M$ is equivalent to a nowhere vanishing section of $\wedge^nT^*M$. This holomorphic $2$-form $\omega$ determines the trivialisation via $$\eta=f\cdot\omega$$ where $f\in H^0(X,\mathcal{O}_X)$ and $\eta\in H^0(X,K_X)$. It would be a good exercise for you to show that the converse is true: if $K_X\cong\mathcal{O}_X$, then such a $2$-form exists.