First Bianchi identity

On any coordinate chart, write $X=X^i{\mathrm{\partial }}_i$ and $Y=Y^i{\mathrm{\partial }}_i$. Applying [F4] to a local smooth function and using the ordinary product rule, the terms with second derivatives cancel and give $[X,Y]=(X^i{\mathrm{\partial }}_i{Y}^j-Y^i{\mathrm{\partial }}_i{X}^j){\mathrm{\partial }}_j\mathrm{.}$ The displayed coefficients are smooth by [F3], so every bracket used below is a smooth vector field.

Expanding the three curvature terms by [F1] and collecting derivatives with the same outer field gives the cyclic sum as $\begin{array}{rl}{\displaystyle }& {\displaystyle {\mathrm{\nabla }}_X({\mathrm{\nabla }}_{Y}Z-{\mathrm{\nabla }}_{Z}Y)+{\mathrm{\nabla }}_Y({\mathrm{\nabla }}_{Z}X-{\mathrm{\nabla }}_{X}Z)+{\mathrm{\nabla }}_Z({\mathrm{\nabla }}_{X}Y-{\mathrm{\nabla }}_{Y}X)}\\ {\displaystyle }& {\displaystyle -{\mathrm{\nabla }}_{[X,Y]}Z-{\mathrm{\nabla }}_{[Y,Z]}X-{\mathrm{\nabla }}_{[Z,X]}Y\mathrm{.}}\end{array}$ By [F2] the three parenthesized differences are $[Y,Z]$, $[Z,X]$, and $[X,Y]$.

Acting on an arbitrary local smooth function $f$ and using [F4], expand the twelve resulting third-order compositions in $[X,[Y,Z]]f+[Y,[Z,X]]f+[Z,[X,Y]]f\mathrm{.}$ Each composition occurs once with sign $+$ and once with sign $-$; hence the sum is zero. Since equality of vector fields is local and is detected by their action on smooth functions, the Jacobi identity holds for $X,Y,Z$.

Apply [F2] once more to pair each term ${\mathrm{\nabla }}_X[Y,Z]$ with $-{\mathrm{\nabla }}_{[Y,Z]}X$, and cyclically. The expression from step 1.2 becomes $[X,[Y,Z]]+[Y,[Z,X]]+[Z,[X,Y]]$, which is zero by step 2.1.