Let $\mathbf{X} \sim N_p(\mu, \Sigma)$ be a multivariate Normal random variable, and $Y \sim \chi^2_{\nu}$ be a $\chi^2$ random variable with $\nu$ degrees of freedom. Then the random variable $\mathbf{X}/\sqrt{Y/\nu}$ follows a Multivariate $t$ distribution with $\nu$ degrees of freedom. The pdf of this distribution is
$$f(\mathbf{x}) = \dfrac{\Gamma((\nu+p)/2)}{\Gamma(\nu/2) \nu^{p/2} \pi^{p/2} |\Sigma|^{1/2} } \left[1 + \dfrac{(\mathbf{x} - \mu)^T \Sigma^{-1} (\mathbf{x}-\mu)}{\nu} \right]^{-(v+p)/2} \,.$$
When $\Sigma$ is a diagonal matrix, the components of $\mathbf{X}$ are independent normal, but note that the pdf of the resulting multivariate $t$ distribution does not decompose into the product of the marginals, since the pdf is:
$$f(\mathbf{x}) = \dfrac{\Gamma((\nu+p)/2)}{\Gamma(\nu/2) \nu^{p/2} \pi^{p/2} |\Sigma|^{1/2} } \left[1 + \dfrac{1}{\nu}\sum_{i=1}^{p}\dfrac{(\mathbf{x} - \mu)^2 }{\sigma^2_i} \right]^{-(v+p)/2} \,.$$
Thus, independence of the underlying normal random variables, does not imply independence of the resulting $t$ variables. Also not that, the since the $\Sigma$ matrix is a diagonal matrix, the $t$ marginal variables are uncorrelated (but not independent).
So the joint distribution of uncorrelated $t$-random variables is above. As for the joint distribution of uncorrelated and independent $t$ random variables, that would just be the product of marginals.
As for using separate $\chi^2$ distributions, for a diagonal $\Sigma$ matrix that should correspond to independent univariate $t$ distributions. I am not quite sure what happens for a non-diagonal $\Sigma$ in this case.
