formula of square of the covariant derivative

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I am stuck with the calculation of $(\nabla ^2 \beta)(X,Y,Z_1,\dots,Z_r)$.

In the following, capital letters are arbitrary vector fields.

Suppose $\beta$ is an $(r,0)$ tensor. Denote $(\nabla \beta)(Y,Z_1,\dots,Z_r) = (\nabla_Y \beta) (Z_1,\dots,Z_r)$, just two ways to express the same thing. Recall the definition

$(\nabla _Y \beta)(X_1,\dots,X_r) := Y[\beta(X_1,\dots,X_r)] - \beta(\nabla_Y X_1,\dots, X_r) - \dots - \beta(X_1,\dots,\nabla_Y X_r) $

Now I try to compute $(\nabla ^2 \beta)(X,Y,Z_1,\dots,Z_r)$

$(\nabla ^2 \beta)(X,Y,Z_1,\dots,Z_r) = (\nabla (\nabla \beta))(X,Y,Z_1,\dots,Z_r)= (\nabla_X (\nabla \beta))(Y,Z_1,\dots,Z_r) = X[ (\nabla \beta)(Y,Z_1,\dots Z_r)] - (\nabla \beta)(\nabla _X Y,Z_1,\dots,Z_r) - (\nabla \beta)(Y,\nabla _X Z_1,\dots,Z_r) - \dots - (\nabla \beta)(Y, Z_1,\dots,\nabla _X Z_r) $

Well, according to Chow's Hamilton Ricci Flow it is true that

$(\nabla ^2 \beta)(X,Y,Z_1,\dots,Z_r) = X[ (\nabla \beta)(Y,Z_1,\dots Z_r)] - (\nabla \beta)(\nabla _X Y,Z_1,\dots,Z_r)$

so I assume that we must have

$ (\nabla \beta)(Y,\nabla _X Z_1,\dots,Z_r) + \dots + (\nabla \beta)(Y, Z_1,\dots,\nabla _X Z_r) = 0 $

and I have no idea how to make the last computation. Thank you for any help.

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Your computation is correct. In the book by Chow, Lu and Ni, it is said

$$(\nabla ^2 \beta)(X,Y,Z_1,\dots,Z_r) = \nabla_X (\nabla \beta)(Y,Z_1,\dots Z_r) - (\nabla_{\nabla _X Y} \beta)(Z_1,\dots,Z_r)$$ which is different from

$$(\nabla ^2 \beta)(X,Y,Z_1,\dots,Z_r) = X[ (\nabla \beta)(Y,Z_1,\dots Z_r)] - (\nabla \beta)(\nabla _X Y,Z_1,\dots,Z_r).$$ Note that

$$(\nabla \beta)(\nabla _X Y,Z_1,\dots,Z_r)\ne (\nabla_{\nabla _X Y} \beta)(Z_1,\dots,Z_r).$$