When there is no torsion the "gammas" are symmetric in the covariant indices
$$\Gamma_{h\,\,\,k}^{\,\,\,j}=\Gamma_{k\,\,\,h}^{\,\,\,j}$$
That means that the term
$$\Gamma_{d\,\,\,c}^{\,\,\,b}(\partial_bv^a+\Gamma_{e\,\,\,b}^{\,\,\,a}v^e)$$
is symmetric in $c$ and $d$.
Symmetric terms in $c$ and $d$ will cancel out when we subtract $\nabla_d\nabla_cv^a$ from $\nabla_c\nabla_dv^a$.
The ordinary partial derivatives
$$\partial_d\partial_cv^a=\partial_c\partial_dv^a$$
are also symmetric in $c$ and $d$ since $v^a(x^j)\in C^2$ (see Wikipedia on Symmetry of second derivatives)
Finally, you need to use the product rule (Leibniz rule)
$$\partial_c(\Gamma_{b\,\,\,d}^{\,\,\,a}v^b)=\partial_c(\Gamma_{b\,\,\,d}^{\,\,\,a})v^b+\Gamma_{b\,\,\,d}^{\,\,\,a}\partial_cv^b$$
which means you will have an extra term compared to your calculations
$$\nabla_c\nabla_dv^a=\partial_c\partial_dv^a+\partial_c(\Gamma_{b\,\,\,d}^{\,\,\,a})v^b+\Gamma_{b\,\,\,d}^{\,\,\,a}\partial_cv^b-\Gamma_{d\,\,\,c}^{\,\,\,b}(\dots)+\Gamma_{b\,\,\,c}^{\,\,\,a}\partial_dv^b+\Gamma_{b\,\,\,c}^{\,\,\,a}\Gamma_{e\,\,\,d}^{\,\,\,b}v^e$$
Now we see that the expression $(\Gamma_{b\,\,\,d}^{\,\,\,a}\partial_cv^b+\Gamma_{b\,\,\,c}^{\,\,\,a}\partial_dv^b)$ is obviously symmetric in $c$ and $d$.
So in conclusion
$$\nabla_c\nabla_dv^a=\partial_c(\Gamma_{b\,\,\,d}^{\,\,\,a})v^b+\Gamma_{b\,\,\,c}^{\,\,\,a}\Gamma_{e\,\,\,d}^{\,\,\,b}v^e+\textit{symmetric contributions}$$
Update
The author is collecting symmetric terms under the umbrella "symmetric contributions" to make the practical computations easier. It's a way of being "lazy". But how does it work? Lets find out! Our goal is to figure out
\begin{equation}
(\nabla_c\nabla_d-\nabla_d\nabla_c)v^a {\tag 1}
\end{equation}
This is actually a measure of the symmetry properties- or rather, the lack thereof- of repeated partial covariant differentiation.
An entity is said to be symmetric in a pair of superscripts (or in a pair of subscripts) if an interchange of the indices concerned does not affect the values of the components.
So for instance, the mixed partial derivatives
\begin{equation}
\partial _c\partial _d v^a=\frac{\partial^2 v^a}{\partial x^c \partial x^d}=\frac{\partial^2 v^a}{\partial x^d \partial x^c}=\partial_d\partial_c v^a{\tag 2}
\end{equation}
is an example of an entity where an interchange of the indices $(c,d)$ does not change the values of the components.
Not all of the terms in $\nabla_c\nabla_d v^a$ are symmetric, but those who are will cancel out when we put everything together in the expression $(\nabla_c\nabla_d-\nabla_d\nabla_c)v^a$
In your updated question you have correctly and laboriously calculated two boxed expressions, one for $\nabla_c \nabla_dv^a$ and one for $\nabla_d\nabla_cv^a$. These represent the complete calculation which means you have already done all the work and there is no need to use this "symmetric contribution"-shenanigans to simplify the calculations. But we can use your calculations to illustrate how symmetric terms cancel each other in a pairwise fashion. I will now go through the symmetric terms, one by one.
The first term in your first box is $\partial_c\partial_d v^a$, the first term in your second box is $\partial_d\partial_c v^a$. Upon forming the expression $(\nabla_c\nabla_d-\nabla_d\nabla_c)v^a$ we see that this part gives us the contribution
\begin{equation}
\partial_c \partial_dv^a -\partial_d\partial_cv^a=\partial_c \partial_dv^a -\partial_c\partial_dv^a=0
\tag{3}
\end{equation}
Notice how the "total" symmetric contribution is zero. The third and fourth terms in your first box are $-\Gamma^b_{dc}(\partial_b v^a+\Gamma^a_{eb}v^e)$. The third and fourth terms in your second box are $-\Gamma^b_{cd}(\partial_b v^a+\Gamma^a_{eb}v^e)$. Again, when we form the expression $(\nabla_c\nabla_d-\nabla_d\nabla_c)v^a$ we see that this part gives us the contribution (remember that $\Gamma^b_{cd}=\Gamma^b_{dc}$)
\begin{equation}
-\Gamma^b_{dc}(\partial_b v^a+\Gamma^a_{eb}v^e)-(-\Gamma^b_{cd}(\partial_b v^a+\Gamma^a_{eb}v^e))=0
\tag{4}\end{equation}
Finally, the sixth and seventh terms in your first box are $\Gamma^a_{bd}\partial_cv^b+\Gamma^a_{bc}\partial_dv^b$. The sixth and seventh terms in your second box are $\Gamma^a_{bc}\partial_dv^b+\Gamma^a_{bd}\partial_cv^b$. This gives us the contribution
\begin{equation}
(\Gamma^a_{bd}\partial_cv^b+\Gamma^a_{bc}\partial_dv^b)-(\Gamma^a_{bc}\partial_dv^b+\Gamma^a_{bd}\partial_cv^b)=0\tag{5}
\end{equation}
In conclusion $\partial_c \partial_dv^a$, $-\Gamma^b_{dc}(\partial_b v^a+\Gamma^a_{eb}v^e)$ and $(\Gamma^a_{bd}\partial_cv^b+\Gamma^a_{bc}\partial_dv^b)$ are symmetric in the indices $(c,d)$ and hence cancel out. This is why the mystical "+symmetric contributions" disappear in the final result.
We are left with the second and fifth term in your first box minus the second an the fifth term in your second box
\begin{equation}
(\nabla_c\nabla_d-\nabla_d\nabla_c)v^a=\partial_c\Gamma^a_{bd}v^b+\Gamma^a_{bc}\Gamma^b_{ed}v^e-\partial_d\Gamma^a_{bc}v^b- \Gamma^a_{bd}\Gamma^b_{ec}v^e\tag{6}
\end{equation}
Renaming dummy index $b\leftrightarrow e$ and putting it in the usual order
\begin{equation}
(\nabla_c\nabla_d-\nabla_d\nabla_c)v^a=\left(\partial_c\Gamma^a_{bd}-\partial_d\Gamma^a_{bc}+\Gamma^a_{ec}\Gamma^e_{bd}- \Gamma^a_{ed}\Gamma^e_{bc}\right)v^b\tag{7}
\end{equation}
Best Answer
Suppose for a contradiction that $$\frac{\partial\left(\partial_cA_d\right)}{\partial\left(\partial_\mu A_\nu\right)}=\delta_c^\color{red}{\nu}\delta_d^\color{red}{\mu}.$$ Take $c=\mu=0$ and $d=\nu=2$ in such equation. Then we have $$\frac{\partial\left(\partial_0A_2\right)}{\partial\left(\partial_0 A_2\right)}=\delta_0^\color{red}{2}\delta_2^\color{red}{0}=0.$$ This is a contradiction, because $$\frac{\partial\left(\partial_0A_2\right)}{\partial\left(\partial_0 A_2\right)}=1.$$ Therefore, $$\frac{\partial\left(\partial_cA_d\right)}{\partial\left(\partial_\mu A_\nu\right)}\neq\delta_c^\color{red}{\nu}\delta_d^\color{red}{\mu}.$$ Now, suppose for a contradiction that $$\frac{\partial\left(\partial_cA_d\right)}{\partial\left(\partial_\mu A_\nu\right)}=\delta_\color{blue}{d}^\mu\delta_\color{blue}{c}^\nu.$$ Take $c=\mu=0$ and $d=\nu=2$ in such equation. Then we have $$\frac{\partial\left(\partial_0A_2\right)}{\partial\left(\partial_0 A_2\right)}=\delta_2^\color{blue}{0}\delta_0^\color{blue}{2}=0.$$ This is also a contradiction, because $$\frac{\partial\left(\partial_0A_2\right)}{\partial\left(\partial_0 A_2\right)}=1.$$ Therefore, $$\frac{\partial\left(\partial_cA_d\right)}{\partial\left(\partial_\mu A_\nu\right)}\neq\delta_\color{blue}{d}^\mu\delta_\color{blue}{c}^\nu.$$ It is worth point out that $\delta_c^\color{red}{\nu}\delta_d^\color{red}{\mu}=\delta_\color{blue}{d}^\mu\delta_\color{blue}{c}^\nu$. However, $\delta_{c}^\mu\delta_{d}^\nu\neq\delta_c^\color{red}{\nu}\delta_d^\color{red}{\mu}$.
In conclusion, we must have $$\frac{\partial\left(\partial_cA_d\right)}{\partial\left(\partial_\mu A_\nu\right)}=\delta_c^\mu\delta_d^\nu.$$