The question is not well-formulated (for instance, it's not clear what you mean by "the right chevalley basis", and Chevalley is a proper name). Most important, your third sentence doesn't make sense: "Let $x_\alpha(t)$ be the corresponding root space." The elements here should be in a related matrix group, which gets complicated when there are longer root strings than in special linear types. That's the main problem here, I think, since your definition of $x_\alpha(t)$ for the case of type $C_2$ isn't precise. Here the long simple root $\alpha$ you start with has a root string involving $\alpha + \beta, \alpha +2\beta$. where $\beta$ is the short simple root. So the expression for $x_\alpha(t)$ gets more complicated. (You should specify what sources you are following, since there are some differences in notation and terminology.)
In Chevalley's original 1955 construction of adjoint Chevalley groups, his starting point is certainly a Chevalley basis of the corresponding complex simple Lie algebra (which is not unique but involves delicate sign choices). Once that is in place, the group elements imitate characteristic 0 exponentials of root vectors. But this gets tricky, especially over fields of small characteristic, since the exponentials have summands with $n!$ in the denominator. The Chevalley basis helps to circumvent this problem in arbitrary characteristic.
[Note too that Steinberg in his 1967-68 Yale lectures broadened the framework to allow for arbitrary representations in characteristic 0; here you build the exponentials into the Kostant basis for the universal enveloping algebra. People working in algebraic K-theory subsequently drew more conclusions from Steinberg's formalism in order to study what happens over $\mathbb{Z}$ or other rings of "integers" in local and global fields.]
ADDED: At first I assumed you were working as Chevalley originally did, with the adjoint representation. Here that involves $10 \times 10$ matrices, so direct calculation is lengthy. But apparently your problem arises when the representation used is the 4-dimensional one, in which case you apparently made a computational error. As Paul comments, the $4 \times 4$ matrix $h_\alpha(t)$ is indeed diagonal. Aside from computational glitches, your term "root space" certainly doesn't apply to the matrix $x_\alpha(t)$. But the matrix notation depends on specifying a (faithful) representation of the Lie algebra you begin with. In Steinberg's approach there is a "Chevalley group" attached to each such representation of the given simple Lie algebra.