For a vector field $X$ and a covariant derivative(Levi-Civita connection for a given metric, for simplicity) $\nabla$ let us suppose that $\nabla_a X_b+\nabla_b X_a=0$ holds. Then it is said that $\nabla_c\nabla_a X_b+\nabla_c\nabla_b X_a=0$ holds trivially. But I cannot understand how this second derivative identity holds.... Isn't $\nabla_c\nabla_bX_a=(\nabla_c\nabla_bX)_a$? Then from by calculations regarding Christoffel symbols, \begin{align} \partial_c(\nabla_aX_b)=\nabla_c\nabla_aX_b+\Gamma^f_{cb}\nabla_aX_f\\\partial_c(\nabla_bX_a)=\nabla_c\nabla_bX_a+\Gamma^f_{ca}\nabla_bX_f\\ \end{align}
Since $\partial_c(\nabla_aX_b)=-\partial_c(\nabla_bX_a)$ we can equate the above two equations by - sign. However, $\Gamma^f_{cb}\nabla_aX_f+\Gamma^f_{ca}\nabla_bX_f$ simply doesn't vanish as should be....As a result, $\nabla_c\nabla_aX_b=\nabla_c\nabla_bX_a$ cannot be obtained.... It seems that I am terribly misunderstanding something and it is extremely frustrating.... Could anyone please help me?
The traditional way of using indexes in the notation for covariant derivatives is notoriously confusing.
If you have a vector field $X$, you would write its components as $X^a$, but in presence of a Riemaniinan metric it is customary to identify vectors with co-vectors (= 1-forms), so you can write a condition such as $\nabla_a X_b+\nabla_b X_a=0$ still thinking of $X$ as of a vector field.
The covariant derivative of $X$ with respect to some connection $\nabla$ is denoted by $\nabla X$, and this is a tensor field with two indexes. Up to the identification mentioned above ("the musical isomorphisms"), you can write the components of $\nabla X$ as $\nabla_a X_b$, as you do. The formulas using the Christoffel symbols allow to express these components in terms of a local coordinate patch.
But then the covariant derivative of $\nabla X$ in turn is denoted as $\nabla \nabla X$, and $\nabla_a \nabla_b X_c$ is a confusing but wide-spread manner of denoting the components of this new tensor with three indexes. Really, it would be better to write $(\nabla \nabla X)_{a b c}$, but who would bother :)
Now, $\nabla_a X_b+\nabla_b X_a=0$ means, that you have a sum of two tensor fields $\nabla X$ and $\text{[Swap a and b]}\nabla X$, where the components of the latter field are given by $$ (\text{[Swap a and b]}\nabla X)_{a b} = (\nabla X)_{b a} $$ For any two tensor fields $T$ and $S$, the covariant derivative enjoys the linearity property: $$ \nabla (T + S) = \nabla T + \nabla S $$
Write it in terms of components, and you get:
$$ (\nabla (T + S))_{abc\dots} = (\nabla T)_{abc\dots} + (\nabla S)_{abc\dots} $$
Applying this to our situation, we see
$$ \begin{align} (\nabla (\nabla X + \text{[Swap a and b]}\nabla X))_{c a b} & = (\nabla \nabla X)_{c a b} + (\nabla \text{[Swap a and b]} \nabla X )_{c a b} \\ & = (\nabla \nabla X)_{c a b} + (\nabla \nabla X)_{c b a} \end{align} $$
Sorry for this ad-hoc $\text{[Swap a and b]}$ notation. It is only used to convey the idea.
Appendix: the second covariant derivative of a vector field in terms of the Christoffel symbols.
As the OP requested in chat, I put here a few lines regarding an expression for the second covariant derivative of a vector field in terms of the Christoffel symbols.
I will assume that the following two identities for the components of the covariant derivatives of a co-vector field (1-form) $\omega$ and a vector field $X$ are known: $$ (\nabla \omega)_{a b} = \nabla_a \omega_b = \partial_a \omega_b - \Gamma_{a b}{}^{c} \omega_c \tag{I} $$
and
$$ (\nabla X)_a{}^b = \nabla_a X^b = \partial_a X^b + \Gamma_{a c}{}^{b} X^c \tag{II} $$
where the Einstein summation convention is used.
Now, let us express the second covariant derivative $\nabla \nabla X$, using these formulas.
Since $\nabla X$ is a tensor field, we may think of it as of a finite sum of tensor products of co-vectors and vectors. Thus, we may run our calculation for a tensor of the form $\omega \otimes X$, and then extend our results by linearity.
So, let us imagine for a moment, that $\nabla X \equiv \omega X$.
Notice, that for a tensor product $(\omega X)_a{}^b \equiv \omega_a X^b$.
Then we can write
$$ (\nabla \nabla X)_{a b}{}^{c} = (\nabla (\omega X))_{a b}{}^{c} = ((\nabla \omega)_{a b} X^c + \omega_b (\nabla X)_{a}{}^{c}) $$
where we have used the Leibniz rule.
Substituting the identities $(I)$ and $(II)$ into the above result, we obtain:
$$ \begin{align} (\nabla \nabla X)_{a b}{}^{c} & = (\partial_a \omega_b - \Gamma_{a b}{}^{d} \omega_d) X^c + \omega_b ( \partial_a X^c + \Gamma_{a d}{}^{c} X^d) \\ & = (\partial_a \omega_b) X^c - \Gamma_{a b}{}^{d} (\omega_d X^c) + \omega_b ( \partial_a X^c) + \Gamma_{a d}{}^{c} (\omega_b X^d) \\ & = \partial_a(\omega_b X^c) - \Gamma_{a b}{}^{d}(\omega_d X^c) + \Gamma_{a d}{}^{c} (\omega_b X^d) \end{align} $$
Now, watch the hands: $\omega X$ was representing $\nabla X$ (this is where we "extend by linearity"), so, we can write:
$$ \begin{align} (\nabla \nabla X)_{a b}{}^{c} & = \partial_a(\nabla_b X^c) - \Gamma_{a b}{}^{d}(\nabla_d X^c) + \Gamma_{a d}{}^{c} (\nabla_b X^d) \end{align} $$
But for $\nabla X$ we can substitute the identity $(II)$ again, and continue the calculation!
So,
$$ \begin{align} (\nabla \nabla X)_{a b}{}^{c} & = \partial_a(\partial_b X^c + \Gamma_{b d}{}^{c} X^d) - \Gamma_{a b}{}^{d}(\partial_d X^c + \Gamma_{d e}{}^{c} X^e) + \Gamma_{a d}{}^{c} (\partial_b X^d + \Gamma_{b e}{}^{d} X^e) \\ & = \partial_a \partial_b X^c + (\partial_a \Gamma_{b d}{}^{c}) X^d + \Gamma_{b d}{}^{c} \partial_a X^d \\ & - \Gamma_{a b}{}^{d}(\partial_d X^c) - \Gamma_{a b}{}^{d} \Gamma_{d e}{}^{c} X^e + \Gamma_{a d}{}^{c} (\partial_b X^d) + \Gamma_{a d}{}^{c} \Gamma_{b e}{}^{d} X^e \end{align} $$
Finally, we can present the above result as
$$ \begin{align} (\nabla \nabla X)_{a b}{}^{c} & = \partial_a \partial_b X^c - \Gamma_{a b}{}^{d} \Gamma_{d e}{}^{c} X^e + \Gamma_{a d}{}^{c} \Gamma_{b e}{}^{d} X^e \\ & + (\partial_a \Gamma_{b d}{}^{c}) X^d + \Gamma_{b d}{}^{c} \partial_a X^d - \Gamma_{a b}{}^{d}(\partial_d X^c) X^e + \Gamma_{a d}{}^{c} (\partial_b X^d) \end{align} $$
where one can clearly recognize the "tensorial" and "non-tensorial" parts.
I hope that this helps.