Is a geodesic in the 4d spacetime still a geodesic after projection onto the 3d space?

But what we can observe is the projected path of the object onto the 3d space, which is itself a Riemannian manifold, with the inherited Riemannian metric from the spacetime.

Not true. To get a 3d space, you have to pick a surface of simultaneity. GR doesn't in general have a preferred surface of simultaneity.

Is the projected geodesic onto the 3d space still a geodesic?

No. For example, the earth orbits the sun, and the sun's field is approximately static, so that there is a preferred time-slicing. With this time-slicing, space is nearly flat, and the earth's orbit is an ellipse, which is not a geodesic.

The spatial metric associated to some familly of local observers is the following (in some coordinates system. I'm using signature $\eta = (1, -1, -1, -1)$): $$\tag{1} h_{\mu \nu} = u_{\mu} \, u_{\nu} - g_{\mu \nu}, $$ where $u^{\mu}$ are the components of the 4-velocity of the local observer. Components (1) define a projector: $$\tag{2} h_{\mu \lambda} \, h^{\lambda}_{\; \nu} = h_{\mu \nu}. $$ Also: $h_{\mu \nu} \, u^{\nu} \equiv 0$ and obviously $h_{\mu}^{\; \lambda} \, g_{\lambda \kappa} \, h^{\kappa}_{\; \nu} \equiv h_{\mu \nu}$. The 3D spatial section defined with this lower metric depends on the familly of observers you select.

You could then compute the 3D Riemann tensor on that lower space. The spacelike geodesics of that lower space have nothing to do with the timelike geodesics of the full 4D metric, despite that (1) is a projector.

Notice that the lower metric (1) isn't compatible with the full connection: $$\tag{3} \nabla_{\mu} \, h_{\lambda \kappa} = \nabla_{\mu} (u_{\lambda} \, u_{\kappa}) \ne 0. $$ To define your lower Riemann tensor and also the 3D geodesics, you'll need a new connection $\tilde{\Gamma}_{\mu \nu}^{\lambda}$ from $h_{\mu \nu}$ such that $$\tag{4} \tilde{\nabla}_{\mu} \, h_{\lambda \kappa} = 0. $$