We have the Gauss curvature equation: $$\langle R(V,W)X,Y\rangle = \langle R'(V,W)X,Y\rangle - \langle II(V,X),II(W,Y)\rangle + \langle II(V,Y),II(W,X)\rangle$$ Here $M$ is an immersion in $N$. $\nabla$ and $\nabla'$ are connections of $M$ and $N$ with the corresponding curvature tensor $R$ and $R'$. $II(X,Y)=({\nabla'}_XY)^\perp$ and $V, W, X, Y$ are smooth vector fields of $M$.
Assuming M is a hypersurface in N, then the following is from page 470 of a paper at http://journals.cambridge.org/download.php?file=%2FPEM%2FPEM2_46_02%2FS0013091502000500a.pdf&code=7621dc2c1ae451af0244640216120c97:
$$R(V,W)X = (R'(V,W)X)^T - \langle AV,X \rangle AW + \langle AW,X \rangle AV$$ Here $()^T\in TM$ is the tangent part of the vector field and $AX=-({\nabla'}_XN)$, $N$ being the normal unit vector.
Can someone give me a hint on how to derive the second equation for hypersurface using the Gauss curvature equation on top? I'm not sure how to start since the Gauss curvature equation is in dot product form while the equation I'm trying to prove is an explicit formula for $R(V,W)X$. Possibly assuming a coordinate patch with basis and letting Y in the Gauss equation equal each element in basis to find coefficient of $R(V,W)X$ in coordinate form?