= Solution
In coordinates, write $\nabla_{\partial_i}\partial_\ell=\Gamma^k_{i\ell}\partial_k$. The <Riemannian curvature two-form> is defined by the matrices
$$
R^k{}_{\ell ij}=\partial_i\Gamma^k_{j\ell}-\partial_j\Gamma^k_{i\ell}+\Gamma^k_{im}\Gamma^m_{j\ell}-\Gamma^k_{jm}\Gamma^m_{i\ell},
$$
with repeated indices summed. It assigns to $X,Y$ the fiberwise <endomorphism> $R(X,Y)$ whose entries are $R^k{}_{\ell ij}X^iY^j$. We can prove both coordinate independence and the required intrinsic identity as follows.
Define an operator on vector fields by $C(X,Y)=[\nabla_X,\nabla_Y]-\nabla_{[X,Y]}$. It is alternating in $X,Y$. For a <smooth function> $f$, the identities $[fX,Y]=f[X,Y]-Y(f)X$ and $\nabla_Y(fZ)=Y(f)Z+f\nabla_YZ$ give
$$
C(fX,Y)Z=fC(X,Y)Z;
$$
the two terms involving $Y(f)\nabla_XZ$ cancel. Alternation gives linearity over <smooth functions> in $Y$ as well. Expanding in the final argument yields
$$
C(X,Y)(fZ)=fC(X,Y)Z+\bigl(X(Yf)-Y(Xf)-[X,Y]f\bigr)Z=fC(X,Y)Z.
$$
Thus $C$ is tensorial in all three arguments, defining a global <endomorphism>-valued two-form.
For coordinate fields $\partial_i,\partial_j$, their bracket is zero. Expand $[\nabla_{\partial_i},\nabla_{\partial_j}]Z$ using $Z=Z^\ell\partial_\ell$. The second derivatives of $Z^\ell$ cancel by equality of mixed partials. The terms containing first derivatives of $Z^\ell$ cancel in pairs, and the remaining coefficient of $Z^\ell\partial_k$ is exactly $R^k{}_{\ell ij}$ above. Hence the coordinate definition equals the global tensor $C$, establishing coordinate independence and proving
$$
\boxed{R(X,Y)=[\nabla_X,\nabla_Y]-\nabla_{[X,Y]}.}
$$
This convention for the <Riemann curvature tensor> is valid for any <affine connection>; in particular it applies to the <Levi-Civita connection>.
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