Direct proof of the second Bianchi identity

Using the product rule for differentiation $$ \nabla_{X} R(Y,Z)W + \nabla_{Y} R(Z,X)W + \nabla_{Z} R(X,Y)W \\= \nabla_X(R(Y,Z)W) + \nabla_Y(R(Z,X)W) + \nabla_Z(R(X,Y)W) \\- R(\nabla_X Y,Z)W - R(Y,\nabla_X Z)W \\- R(\nabla_Y Z,X)W - R(Z,\nabla_Y X)W \\- R(\nabla_Z X,Y)W - R(X,\nabla_Z Y)W \\- R(Y,Z)\nabla_X W - R(Z,X)\nabla_Y W - R(X,Y)\nabla_Z W $$ Also $$R(\nabla_X Y,Z) + R(Y,\nabla_X Z) + R(\nabla_Y Z,X) + R(Z,\nabla_Y X) + R(\nabla_Z X,Y) + R(X,\nabla_Z Y) \\= R(\nabla_X Y - \nabla_Y X,Z) + R(\nabla_Y Z - \nabla_Z Y,X) + R(\nabla_Z X - \nabla_X Z,Y) \\= R(\tau(X,Y) + [X,Y],Z) + R(\tau(Y,Z) + [Y,Z],X) + R(\tau(Z,X) + [Z,X],Y) $$ And $$ \nabla_X(R(Y,Z)W) + \nabla_Y(R(Z,X)W) + \nabla_Z(R(X,Y)W) \\= \nabla_X \nabla_Y \nabla_Z W - \nabla_X \nabla_Z \nabla_Y W + \nabla_Y \nabla_Z \nabla_X W - \nabla_Y \nabla_X \nabla_Z W + \nabla_Z \nabla_X \nabla_Y W - \nabla_Z \nabla_Y \nabla_X W \\ - \nabla_X \nabla_{[Y,Z]} W - \nabla_Y \nabla_{[Z,X]} W - \nabla_Z \nabla_{[X,Y]} W \\= \nabla_Y \nabla_Z \nabla_X W - \nabla_Z \nabla_Y \nabla_X W + \nabla_Z \nabla_X \nabla_Y W - \nabla_X \nabla_Z \nabla_Y W + \nabla_X \nabla_Y \nabla_Z W- \nabla_Y \nabla_X \nabla_Z W \\ - \nabla_{[Y,Z]} \nabla_X W - R(X,[Y,Z]) - \nabla_{[X,[Y,Z]]}W - \nabla_{[Z,X]} \nabla_Y W - R(Y,[Z,X]) - \nabla_{[Y,[Z,X]]}W - \nabla_{[X,Y]} \nabla_Z W - R(Z,[X,Y]) - \nabla_{[Z,[X,Y]]}W \\= R(Y,Z)\nabla_X W + R(Z,X)\nabla_Y W + R(X,Y)\nabla_Z W \\ - R(X,[Y,Z])W - R(Y,[Z,X])W - R(Z,[X,Y])W $$ noting that $\nabla_{[X,[Y,Z]]}W + \nabla_{[Y,[Z,X]]}W + \nabla_{[Z,[X,Y]]}W = \nabla_{[X,[Y,Z]] + [Y,[Z,X]] + [Z,[X,Y]]}W = 0$ by the Jacobi identity.

Now combine to get $$ \nabla_{X} R(Y,Z) + \nabla_{Y} R(Z,X) + \nabla_{Z} R(X,Y) = R(X,\tau(Y,Z)) + R(Y,\tau(Z,X)) + R(Z,\tau(X,Y))$$

Note the argument is simpler if one has that all the Lie brackets are zero, which is a much weaker assumption than normal coordinates, etc. I would say that the core of the proof is this identity: $$ \nabla_X \nabla_Y \nabla_Z W - \nabla_X \nabla_Z \nabla_Y W + \nabla_Y \nabla_Z \nabla_X W - \nabla_Y \nabla_X \nabla_Z W + \nabla_Z \nabla_X \nabla_Y W - \nabla_Z \nabla_Y \nabla_X W \\= \nabla_Y \nabla_Z \nabla_X W - \nabla_Z \nabla_Y \nabla_X W + \nabla_Z \nabla_X \nabla_Y W - \nabla_X \nabla_Z \nabla_Y W + \nabla_X \nabla_Y \nabla_Z W- \nabla_Y \nabla_X \nabla_Z W $$

Finally let me comment that the problem with using normal coordinates is that you implicitly use that the connection comes from a Riemannian metric. The proofs given here are simple manipulations of the definitions.

You can find such a proof in my book Riemannian Manifolds: An Introduction to Curvature, pp. 123-124. The computation is hugely simplified by choosing a point and then extending the given vectors to vector fields whose brackets and first covariant derivatives both vanish at the given point. (This can be done, for example, by taking them to be coordinate vectors in Riemannian normal coordinates, but that's not the only way to do it.) From there, it's just a matter of writing down the definition of the covariant derivative of $R$ three times with indices cyclically permuted, and adding.

You can do the whole proof without relying on a special frame, just by keeping track of the commutators and first covariant derivatives, and noting that they all cancel. But why would you want to?