To fix notation, we define the Riemann tensor as
$$
\operatorname{Rm}(x,y,z,w) = \langle R(x,y)w, z \rangle,
$$
and the sectional curvature as $K(x,y) = \operatorname{Rm}(x,y,x,y) / (|x|^2|y|^2 - \langle x,y \rangle^2)$.
The symmetries of the curvature tensor $\operatorname{Rm}$ mean that the bilinear form $b$ on $\bigwedge^2 V$ defined by
$$
b(x \wedge y, z \wedge w) = \operatorname{Rm}(x,y,z,w)
$$
is actually well defined, and we note that the sectional curvature of the metric is just the quadratic form defined by $b$ (divided by the square of the norm of $x \wedge y$). Thus the sectional curvature determines the curvature tensor, because a quadratic form determines a unique bilinear form by polarization.
We now pull out of our hat the tensor
$$
\operatorname{Rm}'(x,y,z,w) = \langle x, z \rangle \langle y, w \rangle
- \langle x, w \rangle \langle y, z \rangle
$$
(or recall where the idea of constant sectional curvature comes from and calculate the curvature tensor of the sphere). This tensor has sectional curvature
$$
K'(x,y)
= \frac{|x|^2 |y|^2 - \langle x, y \rangle^2}{|x|^2|y|^2 - \langle x, y \rangle^2} = 1.
$$
It thus follows that if our curvature tensor $R$ has constant sectional curvature $C$, then $R = C R'$.
We can now stare at $\operatorname{Rm}'$ for a couple of minutes and see that we must have $R'(x,y)z = \langle y, z \rangle \, x - \langle x, z \rangle \, y$, because that is a tensor such that
$$
\langle R'(x,y)w, z \rangle
= \operatorname{Rm}'(x,y,z,w)
$$
for all $w$.
I did in fact see that thread when trying to work out how to solve this issue. But felt that the fact there was no concrete answer there and that it was 6 months old a new thread was fine!
– user291678 Apr 08 '17 at 09:57