0. Caveat Lector: This was done before I drank my morning coffee, so there may be some errors in the reasoning (well, the physical reasoning, the mathematics should be kosher).
1. Perfect Fluid. So we have two stress-energy tensors here. One is the stress energy tensor for a perfect fluid
$$\tag{1}T^{\alpha\beta}_{\text{fluid}} = \rho \, u^\alpha \, u^\beta + p \, h^{\alpha\beta}$$
where we have
- the worldlines of the fluid's particles have velocity $u^\alpha$
- the projection tensor $h_{\alpha\beta} = g_{\alpha\beta} + u_\alpha \, u_\beta$ projects other tensors onto hyperplane elements orthogonal to $u^\alpha$
- the matter density is given by the scalar function $\rho$,
- the pressure is given by the scalar function $p$.
We'd need extra terms if there were heat flow or shear involved.
2. Scalar Field. Now, we have another distinct stress-energy tensor for a massless scalar field:
$$\tag{2}T^{\mu \nu}_{\text{scalar}} =\partial^{\mu}\phi\, \partial^{\nu}\phi-\frac{1}{2}g^{\mu \nu}\partial_{\rho}\phi\,\partial^{\rho}\phi$$
We would use this equation when modeling, e.g., massless pions (or some other massless spin-0 field).
3. Problem: Are these two related?
Now if we take our matter density to be, in the appropriate units,
$$\tag{3a} \rho = 1 + \frac{1}{2}\partial_{\rho}\phi\,\partial^{\rho}\phi $$
and the pressure
$$\tag{3b} p = \frac{-1}{2}\partial_{\rho}\phi\,\partial^{\rho}\phi $$
then (2) resembles (1). This is after pretending $\partial^{\mu}\phi=u^{\mu}$, which terrifies the original poster (but that's what condensed matter physicists do, so I suppose I could end here content).
Is this kosher?
We should first note if we wanted to take the derivative of some function along the worldline $x^{\mu}(s)$ with respect to the "proper time" (length) $s$ we have
$$\tag{4} \frac{\mathrm{d}f}{\mathrm{d}s}=\frac{\mathrm{d}x^{\mu}}{\mathrm{d}s}\frac{\partial f}{\partial x^{\mu}}$$
by the chain rule. For general relativity, we use the "comma-goes-to-semicolon" rule, but for a scalar quantity $f$ we have
$$ \nabla_{\mu}f = \partial_{\mu}f.$$
(If this is not obvious, the reader should consider it an exercise to prove it to him or herself.) The punchline: identifying $\partial^{\mu}\phi=u^{\mu}$ is kosher. How?
Observe in Equation (4) the guy in front, the $\mathrm{d}x^{\mu}/\mathrm{d}s$ is just some vector. So in the very, very special case that equations (3a) and (3b) hold, and $\mathrm{d}x^{\mu}/\mathrm{d}s=(1,0,0,0)$, we see that we can indeed recover the first stress-energy tensor as a special case of the scalar field's stress-energy tensor.