Use $g_{\alpha\beta}$ to compute $R_{\alpha\beta\mu\nu}$.
Use $R_{\alpha\beta\mu\nu}$ to compute $R_{\alpha\beta}$ and $R$.
Use $R_{\alpha\beta}$ and $R$ to compute $G_{\alpha\beta}$.
Use $G_{\alpha\beta}$ to compute $T_{\alpha\beta}$.
Use an unknown $\rho$ and $p$ to express the stress tensor in terms of $U_0,$ $U_1,$ $U_2,$ $U_3,$ $\rho,$ $p,$ and the metric.
You have six unknowns, the trace $T=T^\alpha_{\,\,\,\beta}$ tells you how $\rho$ and $p$ are related. So write everything in terms of
$U_0,$ $U_1,$ $U_2,$ $U_3,$ $\rho,$ $T,$ and the metric. So only five unknowns left.
You can look at $T_{00}$ to learn $U_{0}$ in terms of $T_{00},$ $\rho,$ $T,$ and the metric.
You can look at $T_{11}$ to learn $U_{1}$ in terms of $T_{11},$ $\rho,$ $T,$ and the metric.
You can look at $T_{22}$ to learn $U_{2}$ in terms of $T_{22},$ $\rho,$ $T,$ and the metric.
You can look at $T_{33}$ to learn $U_{3}$ in terms of $T_{33},$ $\rho,$ $T,$ and the metric.
So now, since we know $T$ and the metric, there is really only one unknown, $\rho$.
Now. If $\rho$ and $T$ are both zero, it is all hopeless since the fluid has no gravitational effects (it has no energy and no pressure) so could have any four velocity whatsoever and you wouldn't know it. If $\rho$ is nonzero then you can consider that it was $\rho$ and $p/\rho$ that were the initial unknowns you didn't care about. And $\rho$ just affects the overall size of the stress energy tensor, so when it gets bigger (for fixed $p/\rho$) every component of the stress energy tensor just gets bigger by the same factor.
So you can't always find a four velocity from the metric, but hopefully you can find it in your situation. Don't forget that the four velocity is a unit vector, so you can (using the metric) to get an equation relating the components to each other, hence get another equation with the density $\rho.$