I) In Palatini $f(R)$ gravity, the Lagrangian density is$^1$
$$ {\cal L}(g,\Gamma)~=~ \frac{1}{2\kappa}\sqrt{-g} f(R) + {\cal L}_{\rm m}; \tag{1}$$
with matter Lagrangian density ${\cal L}_{\rm m}$;
with scalar curvature
$$R~:=~ g^{\mu\nu} R_{\mu\nu}(\Gamma);\tag{2}$$
with Ricci curvature $R_{\mu\nu}(\Gamma)$; and where
$$\Gamma^{\lambda}_{\mu\nu}~=~\Gamma^{\lambda}_{\nu\mu}\tag{3}$$
is an arbitrary torsionfree$^2$ connection.
II) As OP mentions, the word Palatini refers to that the metric $g_{\mu\nu}$ and the connection $\Gamma^{\lambda}_{\mu\nu}$ are independent variables$^3$. We therefore get two types of EL equations:
The EL equations
$$ f^{\prime}(R)R_{\mu\nu} -\frac{1}{2}f(R)g_{\mu\nu}~\stackrel{(1)+(5)}{\approx}~\kappa T_{\mu\nu} \tag{4}$$
for the metric $g_{\mu\nu}$ are the generalization of EFE, where
$$T^{\mu\nu}~:=~\frac{2}{\sqrt{-g}}\frac{\delta S_{\rm m}}{\delta g_{\mu\nu}} \tag{5} $$
is the matter Hilbert stress-energy-momentum (SEM) tensor.
[In eq. (4) the $\approx$ symbol means equality modulo equations of motion. In this answer, we use $(-,+,\ldots,+)$ Minkowski sign convention in $d$ spacetime dimensions.]
If the matter action $S_{\rm m}$ doesn't depend on the connection $\Gamma^{\lambda}_{\mu\nu}$, then the EL equations
$$ \nabla_{\lambda}\hat{\mathfrak{g}}^{\mu\nu}
~\stackrel{(1)}{\approx}~0,\qquad
\hat{\mathfrak{g}}^{\mu\nu}
~:=~\sqrt{-g} f^{\prime}(R) g^{\mu\nu}
~\stackrel{(8)}{=}~\sqrt{-\hat{g}} \hat{g}^{\mu\nu}, \tag{6}$$
for the connection $\Gamma^{\lambda}_{\mu\nu}$ turn out to be the metric compatibility condition
$$\nabla_{\lambda}\hat{g}_{\mu\nu} ~\stackrel{(6)+(8)}{\approx}~0\tag{7}$$
for a conformally equivalent metric
$$ \hat{g}_{\mu\nu}~:=~f^{\prime}(R)^{\frac{2}{d-2}} g_{\mu\nu}, \tag{8}$$
known as the Einstein frame metric.
In other words, the classical solution for $\Gamma^{\lambda}_{\mu\nu}$ is the Levi-Civita connection for the Einstein frame metric $\hat{g}_{\mu\nu}$.
III) So Einstein gravity (GR) with a possible cosmological constant
$$ f(R)~=~R-2\Lambda,\tag{9} $$
or equivalently
$$ f^{\prime}(R)~=~1,\tag{10}$$
corresponds to the special case where the two metrics $g_{\mu\nu}$ and $\hat{g}_{\mu\nu}$ coincide, and hence $\Gamma^{\lambda}_{\mu\nu}$ becomes the Levi-Civita connection for $g_{\mu\nu}$.
--
$^1$ It is natural to replace the Lagrangian density (1) with the extended Lagrangian density
$$ \tilde{\cal L}(g,\Gamma,\Phi)~=~ \frac{1}{2\kappa}\sqrt{-g}\{\Phi R-V(\Phi)\} + {\cal L}_{\rm m}; \tag{11}$$
with auxiliary scalar dilaton field $\Phi$; and where the potential
$$V(\Phi)~:=~\sup_r(\Phi r -f(r))\tag{12}$$
is the Legendre transform of the function $f$. If we integrate out the auxiliary scalar field $\Phi$, we then return to the $f(R)$ Lagrangian density (1) which we started from! The EL equations
$$ \nabla_{\lambda}\hat{\mathfrak{g}}^{\mu\nu}
~\stackrel{(1)}{\approx}~0,\qquad
\hat{\mathfrak{g}}^{\mu\nu}
~:=~\sqrt{-g} \Phi g^{\mu\nu}
~\stackrel{(14)}{=}~\sqrt{-\hat{g}} \hat{g}^{\mu\nu}, \tag{13}$$
for the connection $\Gamma^{\lambda}_{\mu\nu}$ become the metric compatibility condition (7) for the Einstein frame metric
$$ \hat{g}_{\mu\nu}~:=~\Phi^{\frac{2}{d-2}} g_{\mu\nu}. \tag{14}$$
After the connection $\Gamma^{\lambda}_{\mu\nu}$ has been integrated out, the Lagrangian density (11) becomes
$$ \tilde{\cal L}(g,\Phi)~=~ \frac{1}{2\kappa}\sqrt{-\hat{g}}\{R(\hat{g})-\Phi^{\frac{d}{2-d}}V(\Phi)\} + {\cal L}_{\rm m} , \tag{15}$$
where eq. (14) is implicitly assumed.
However, unfortunately the Legendre transform $V$ does not exist for Einstein gravity (9), so we will not consider the extended Lagrangian density (11) further in this answer.
$^2$ One could allow a non-dynamical torsion piece, but we will not pursue this here for simplicity. For more on torsion, see e.g. also this Phys.SE post.
$^3$ Normally in non-Palatini formulations, we integrate out the connection $\Gamma^{\lambda}_{\mu\nu}$ and keep the metric $g_{\mu\nu}$. Eddington & Schrödinger proposed the opposite! Let us analyze this possibility here. Define for later convenience a double-index notation $M=\mu\mu^{\prime}$ and the following shorthand notation
$$\frac{f(R)}{2f^{\prime}(R)}~=:~\hat{f}(R) ~\equiv~ \hat{f}_0+\hat{f}_1R +\hat{f}_2(R).\tag{16}$$
Let us only consider the vacuum
$$T_{\mu\nu}~=~0.\tag{17}$$
from now on. We then have
$$g^M~\stackrel{(4)+(16)+(17)}{\approx}~\hat{f}(R)R^M,\tag{18}$$
where $R^M$ is the inverse Ricci tensor. Equivalently, we have
$$\left(\delta^M_N - \hat{f}_1 R^M R_N\right) g^N~\stackrel{(16)+(18)}{\approx}~\left(\hat{f}_0+f_2(R) \right)R^M.\tag{19}$$
So we get a fixed-point equation for the inverse metric
$$g^N~\stackrel{(19)}{\approx}~\left(\delta^N_M +\frac{\hat{f}_1}{1-d\hat{f}_1} R^N R_M\right)\left(\hat{f}_0+\hat{f}_2(R) \right)R^M$$
$$~=~ \frac{1}{1-d\hat{f}_1}\left(\hat{f}_0+\hat{f}_2\left(g^MR_M\right) \right)R^N.\tag{20}$$
Let us specialize to Einstein gravity (9). Then
$$\hat{f}_0~=~-\Lambda;\qquad\hat{f}_1~=~\frac{1}{2};\qquad\hat{f}_2(R)~=~0.\tag{21}$$
The inverse metric becomes
$$g^N~\stackrel{(20)+(21)}{\approx}~\frac{2\Lambda}{d-2}R^N.\tag{22}$$
And hence
$$R~\stackrel{(22)}{\approx}~\frac{2d}{d-2}\Lambda,\tag{23}$$
and
$$g_{\mu\nu}~\stackrel{(2)+(22)}{\approx}~\frac{d-2}{2\Lambda}R_{\mu\nu}(\Gamma).\tag{24}$$
So the EH Lagrangian density becomes Born-Infeld-like:
$${\cal L}(\Gamma)~\stackrel{(1)+(17)+(23)+(24)}{\approx}~\frac{1}{\kappa}\left(\frac{d-2}{2\Lambda}\right)^{\frac{d}{2}-1}\sqrt{-\det(R_{\mu\nu}(\Gamma))}.\tag{25}$$
Note that the Eddington-Schrödinger action (25) only works for a non-zero cosmological constant $\Lambda\neq 0$.
Best Answer
OP is right that Wald assumes that all connections are torsionfree, cf. point 5 on p. 31. Hence $C^c_{\ ab} = C^c_{\ ba}$.
As to why the terms inside the parentheses in eq. (E.1.21) need to be symmetrized over $a$ and $b$, consider the following analogy:
If $M^{ab}S_{ab}=0$ for all symmetric matrices $S_{ab}$, then we can only conclude that $M^{ab}=-M^{ba}$ is an antisymmetric matrix.