Second Covariant Derivative Leibniz Rule I’ve been trying to understand why


  $$\nabla_{u,v}^2 w := (\nabla(\nabla w) )(u, v) = \nabla_u{(\nabla_v w)}-\nabla_{\nabla_u v}w$$


Should be true. Is there some easy way to prove this identity? I'd like to see a proof that doesn't expand out the product in terms of it's coordinates if possible as well. Thanks in advanced!
Remember that we define the covariant derivative of arbitrary tensors so that it satisfies the Leibniz rule and commutes with contractions.

The term $\nabla W(V)$ is the contraction of $\nabla W\otimes V.$ The Leibniz rule tells us that 
$$\nabla_U (\nabla W \otimes V) = (\nabla_U(\nabla W)) \otimes V +\nabla W\otimes\nabla_UV.$$

Contracting this equation yields

$$\nabla_U (\nabla W (V)) = (\nabla_U(\nabla W))(V) +(\nabla W)(\nabla_UV).$$

Recalling the definition $(\nabla W)(X) = \nabla_X W,$ we can rewrite this as the desired

$$ \nabla_U(\nabla_V W) = (\nabla (\nabla W))(U,V)+ \nabla_{\nabla_U V} W.$$
Use the semilatus theorem that states $\delta^i_j=\delta^j_i$ if and only if $i=j$, followed by $$\Gamma_i\oplus \Gamma_j=\sum_{i=1}^{k}\hat{\Gamma}_i$$ where $\hat{\Gamma}$ is a half-metric tensor under a morphic dilational phase. If we use this on gradient fields, we have $$\nabla \Gamma_i\oplus \Gamma_j=\sum_{i=1}^{k}\nabla_i \hat{\Gamma}_i$$
Hence, use the same for your argument. If you combine each psuedoterm correctly, you'll get $$\nabla_{u,v}^2 w := (\nabla(\nabla w) )(u, v) = \nabla_u{(\nabla_v w)}-\nabla_{\nabla_u v}w$$ as you wanted to prove! 

That being said, there are alternative methods. This is the method I learnt when I did tensor field theory. Hope this helps!