Setting aside the assumption that $\phi$ be a polynomial mapping for the moment, if one makes the 'nondegeneracy' assumptions
 
 1. $\mathrm{dim}\ \mathrm{span}\bigl( \phi_s(s,t), \phi_t(s,t)\bigr) =2 $,
 2. $\mathrm{dim}\ \mathrm{span}\bigl( \phi_s(s,t), \phi_t(s,t), \phi_{ss}(s,t), \phi_{st}(s,t), \phi_{tt}(s,t) \bigr) = 3$ for all $(s,t)\in[0,1]^2$, and 
 3. the subspace $W(s,t) =\mathrm{span}\bigl( \phi_s(s,t), \phi_t(s,t), \phi_{ss}(s,t), \phi_{st}(s,t), \phi_{tt}(s,t) \bigr)\subset\mathbb{R}^4$ is not constant, in the sense that $W:[0,1]^2\to \mathrm{Gr}_3(\mathbb{R}^4)\simeq\mathbb{RP}^3$ has nonvanishing differential,

then one can show that the surface $\phi\bigl([0,1]^2\bigr)\subset\mathbb{R}^4$ is ruled (and does not lie in an affine $3$-space).  

Such surfaces locally depend on three arbitrary functions of one variable in Cartan's sense.  One way of describing them is this:  Let $\Lambda$ be the space of (affine) lines in $\mathbb{R}^4$, a manifold of dimension $6$.  Consider the ($9$-dimensional) bundle $\pi:F\to\Lambda$ whose fiber over $\lambda\in\Lambda$ is the flag variety of the $3$-dimensional vector space $\mathbb{R}^4/\lambda'$, where $\lambda'\subset\mathbb{R}^4$ is the linear subspace parallel to $\lambda$. Then there exists a smooth $4$-plane field $D\subset TF$ such that, if $\gamma\subset F$ is a generic curve tangent to $D$, then regarding $\gamma$ as a $1$-parameter family of affine lines via $\pi(\gamma)\subset \Lambda$, the union of these lines is a surface satisfying the above assumptions.  (Here 'generic' means that the tangents to $\gamma$ do not lie in triple of hyperplanes in $D$.)

I don't know, offhand, whether there are any polynomial solutions to these conditions, but I think that shouldn't be hard to answer.  (I suspect that there are such polynomial solutions, but I haven't had time to check this.)

If you are interested, I can put in the analysis that yields these results.