Let $M$ be an $n-m$ dimensional sub-manifold of $\mathbf R^n$ defined by the following set of equations: \begin{equation} f_1(\vec x)=0, \\ \vdots \\ f_m(\vec x)=0, \end{equation} where $\vec x$ are coordinates in $\mathbf R^n$.
Given a point $\vec x_0$ in $M$, how can the Ricci scalar (calculated from the induced metric on $M$) at $\vec x_0$ be expressed in terms of the functions $f_1,...,f_m$?
This is an extended comment answering Ricci scalar of sub-manifold of $\mathbf R^n$
Assuming the $f_i$ are independent, at every point $x$ their gradients span the orthogonal complement to the tangent space of the manifold.
The second fundamental form is given by the normal projection of the ambient derivative. So if $V,W$ are tangent vector fields to $M$, then at the point $p$ you have that $II_p(V,W)$ is the projection to the span of $\{\nabla f_i\}$ of $\partial_V W$. Using that the $f_i$ are independent, this means $II_p(V,W)$ is uniquely determined by the $m$ values $$ \langle \partial_V W, \nabla f_i(p) \rangle $$ (The reconstitution formula, when $\nabla f_i$ are orthogonal, is just to multiply this against $|\nabla f_i(p)|^{-2} \nabla f_i(p)$ and sum. When they are not, there's a matrix inversion involved which makes the formula messy.)
The above inner product, using that $\langle W,\nabla f_i(p)\rangle = 0$ as $W$ is a tangent and $\nabla f_i(p)$ is a normal, can be rewritten as $ \langle W, \partial_v \nabla f_i(p)\rangle$ which is nothing more than the Hessian of $f_i$ evaluated in the direction $V\otimes W$.
More about the matrix inversion: the idea is that given any vector in the normal space you can write it as $$ \sum \alpha_i \nabla f_i $$ You wish to reconstruct $\alpha_i$ from the measuraments $$ \beta_j = \langle\sum_i \alpha_i \nabla f_i, \nabla f_j\rangle = \sum_i \underbrace{\langle \nabla f_j , \nabla f_i \rangle}_{= A_{ji}} \alpha_i $$ Linear independence guarantees that the matrix $A_{ji}$ is invertible; in fact you have explicit formulas of this in terms of cofactors. These are highly multilinear expressions in terms of $\nabla f$, but do not involve Hessians.
For now, denote by $B_{ji}$ the inverse matrix to $A_{ji}$ above. So you can write $$ II_p(V,W) = \sum_{i,j} B_{ji} \nabla f_j \nabla^2_{V,W} f_i $$
Note finally for the expression that shows up in the Gauss equations $$ \langle II_p(V,W), II_p(X,Y)\rangle = \sum_{i,j,k,l} \langle B_{ji} \nabla f_j \nabla^2_{V,W} f_i , B_{kl} \nabla f_k \nabla^2_{X,Y} f_l\rangle $$ the RHS can be reorganized as $$ \sum_{i,j,k,l} B_{ji} B_{kl} A_{jk} \nabla^2_{V,W} f_i \nabla^2_{X,Y} f_l = \sum_{i,l} B_{li} \nabla^2_{V,W} f_i \nabla^2_{X,Y} f_l$$
So the formula for the Riemann curvature (with all indices lowered) is actually not too bad.
For the Ricci scalar you also need to take the trace, which requires computing the tangential projection using $\nabla f_i$: you can do this just by subtracting the normal projection from the identity. This can complicate the expression quite a bit.
Thanks Willie but I find your answer slightly hard to follow. The following is what I uncovered.
Given a basis $\{e_a\}$ (with $a=1,...,n-m$) for the tangent space $TM$ the second fundamental form, $\alpha(\cdot,\cdot)$, in components, is given by \begin{equation} \alpha(e_a,e_b)\equiv \alpha_{ab}^\mu\partial_\mu=(e^\nu_a\partial_\nu e^{\mu}_b-g^{cd}e_{c}^{\alpha}e^{\mu}_d e^\nu_a\partial_v e^\alpha_b)\partial_\mu \end{equation} where $g^{cd}$ is the inverse of the matrix $g_{cd}=e_c^\mu e_d^\mu$. From this the Ricci scalar can be computed using the Gauss formula.
The tangent space $T_{\vec x}M$ is equal to the null-space of the matrix $\frac{\partial f_i(\vec x)}{\partial x^\mu}$. So by finding a basis for this null space, one can find the vectors $\{e_a\}$ and therefore the Ricci scalar.