There is no "universally correct" definition, but $K^0(X)$ usually denotes the K-theory of the exact category of vector bundles over $X$. As you say, there are various approaches for (higher) algebraic K-theory (summarized in Weibel's K-book, for instance), and which one one uses depends on the applications, but also on the realizability. Namely, there are lots of exact sequences of vector bunbles on projective varieties which give you relations in the K-theory, but these exact sequences don't split. Although this makes $K^0(X)$ relatively small compared to $K^0_{\oplus}(X)$ (the K-theory of the monoidal category of vector bundles w.r.t. $\oplus$) and perhaps doesn't contain enough information for what you want, it is at least computable! For example, $K^0(\mathbb{P}^r) \cong \mathbb{Z}[x]/(x^{r+1})$ is well-known (SGA 6), but little is known about $K^0_{\oplus}(\mathbb{P}^r)$ (cf. MO/20444): It is the free abelian group on indecomposable vector bundles on $\mathbb{P}^r$ (Atiyah), but these have not been classified yet.

Grothendieck's celebrated and important resolution theorem won't tell you anything about $K^0_{\oplus}(X)$ because of the appearance of exact seqeunces and alike, so there won't be a nice comparision between K-theory and G-theory. Many other theorems for $K^0$, such as Grothendieck's homotopy-invariance $K^0(X) \cong K^0(X \times \mathbb{A}^1)$ will probably fail for $K^0_{\oplus}$. The upshot is: Although $K^0_{\oplus}$ contains more information, it is not flexible enough for computations.

If $X$ is an affine scheme, the canonical map $K^0_{\oplus}(X) \to K^0(X)$ is an isomorphism. Because of that, many authors define $K_0(R)$ of a ring $R$ to be the free abelian group on iso-classes of f.g. proj. $R$-modules modulo the relation $[P \oplus Q] = [P]+[Q]$, but this should not be seen as the correct definition for the general case of a scheme. In particular I don't agree with your sentence "... the monoidal category structure which seems more natural since that is how for instance K-Theory of rings is defined."