Curvature is obtained by commuting two covariant derivatives without a bracket correction

Using [F1] but no bracket term gives $C(X,fY)Z={\mathrm{\nabla }}_X(f{\mathrm{\nabla }}_{Y}Z)-f{\mathrm{\nabla }}_Y{\mathrm{\nabla }}_{X}Z=fC(X,Y)Z+X(f){\mathrm{\nabla }}_{Y}Z\mathrm{.}$ The extra derivative term shows the precise obstruction to tensoriality.

For an explicit witness, take $M=\mathbb{R}$ with its flat affine connection ${\mathrm{\nabla }}_{a{\mathrm{\partial }}_x}(b{\mathrm{\partial }}_x)=a{b}^{\mathrm{\prime }}{\mathrm{\partial }}_x$, and put $X=Y={\mathrm{\partial }}_x$, $f(x)=x$, and $Z=x{\mathrm{\partial }}_x$. Then $C(X,X)Z=0$, whereas direct differentiation gives $C(X,fX)Z={\mathrm{\nabla }}_X(x{\mathrm{\partial }}_x)-{\mathrm{\nabla }}_{xX}({\mathrm{\partial }}_x)={\mathrm{\partial }}_x\mathrm{\ne }0=fC(X,X)Z\mathrm{.}$ Thus the raw commutator fails function-linearity even in dimension one with a flat torsion-free connection.

By [F2], ${\mathrm{\nabla }}_{[X,fY]}Z=f{\mathrm{\nabla }}_{[X,Y]}Z+X(f){\mathrm{\nabla }}_{Y}Z\mathrm{.}$ Subtracting this expression from step 1.1 cancels the extra term and gives $R(X,fY)Z=fR(X,Y)Z$, as asserted by [F3]. This identifies exactly why the bracket correction cannot be omitted.

The counterexample uses a nonempty boundaryless one-manifold and nonzero $Z$; empty and zero-dimensional manifolds cannot witness the failure because all vector fields vanish there. The calculation is local and remains valid in a boundary chart away from its endpoint. No metric, nondegeneracy, orientation, endpoint limit, or choice axiom is used, and no biconditional is asserted.