Permit me the use of Latin indices instead of Greek indices and the convention $\nabla_a K_b=K_{b;a} $. So we wish to prove
$\newcommand{\Tud}[3]{{#1}^{#2}_{\phantom{#2}{#3}}}$
$$\Tud{K}{a}{;b c} = \Tud{R}{a}{b c d} K^d$$
where
$$\Tud{V}{a}{;b c} - \Tud{V}{a}{;c b} = \Tud{R}{a}{d c b} V^d$$
and
$$K_{a ; b} + K_{b ; a} = 0$$
Differentiating the last equation, we get
$$K_{a ; b c} + K_{b ; a c} = 0$$
so, relabelling and summing,
$$K_{a ; b c} + K_{b ; a c} - K_{b ; c a} - K_{c ; b a} + K_{c ; a b} + K_{a ; c b} = 0$$
hence,
$$K_{a; b c} + K_{a ; c b} = R_{b d a c} K^d + R_{c d a b} K^d$$
By the interchange symmetry $R_{a b c d} = R_{c d a b}$, and raising indices, we get
$$\Tud{K}{a}{;b c} - \Tud{R}{a}{b c d} K^d = -(\Tud{K}{a}{;c b} - \Tud{R}{a}{c b d} K^d)$$
On the other hand, by the first Bianchi identity and antisymmetry, we have
$$\Tud{R}{a}{d c b} = \Tud{R}{a}{b c d} + \Tud{R}{a}{c d b}$$
Hence we get
$$\Tud{K}{a}{;b c} = \Tud{K}{a}{; c b} + \Tud{R}{a}{d c b} K^d = \Tud{K}{a}{;c b} + \Tud{R}{a}{b c d} K^d + \Tud{R}{a}{c d b} K^d$$
and therefore
$$\Tud{K}{a}{;b c} - \Tud{R}{a}{b c d} K^d = \Tud{K}{a}{;c b} - \Tud{R}{a}{c b d} K^d$$
The conclusion follows.
A $(k,l)$-tensor field $A$ on smooth manifold $M$ is a smooth section
$$
A : M \longrightarrow T^{(k,l)}TM,
$$
where $T^{(k,l)}TM = \coprod_{p \in M} \underbrace{T_pM \otimes \cdots \otimes T_pM}_{k \text{ times}} \otimes \underbrace{T^*_pM \otimes \cdots \otimes T^*_pM}_{l \text{ times}} $.
Any tensor field has property that for any smooth vector fields $X_1,\dots,X_l$ and smooth covector fields $\omega^1,\dots,\omega^k$ we have a smooth function
$$
A(\omega^1,\dots,\omega^k,X_1,\dots,X_l) : M \longrightarrow \mathbb{R}
$$
defined as $A(\omega^1,\dots,\omega^k,X_1,\dots,X_l)(p) = A_p(\omega^1|_p,\dots,\omega^k|_p,X_1|_p,\dots,X_l|_p) \in \mathbb{R}$. So $(k,l)$-tensor field $A$ induces a map
$$
\tilde{A} : \underbrace{\mathfrak{X}^*(M)\times\cdots\times\mathfrak{X}^*(M)}_{k \text{ times}} \times \underbrace{\mathfrak{X}(M) \times \cdots \times \mathfrak{X}(M)}_{l \text{ times}} \longrightarrow C^{\infty}(M).
$$
An important fact is that this map is multilinear over $C^{\infty}(M)$. It turns out that this in fact characterize tensor fields.
$\textbf{Tensor Characterization Lemma}$ ( Lee's Introduction to Smooth Manifolds ) A map
\begin{equation}
\tau : \underbrace{\mathfrak{X}^*(M)\times\cdots\times\mathfrak{X}^*(M)}_{k \text{ times}} \times \underbrace{\mathfrak{X}(M) \times \cdots \times \mathfrak{X}(M)}_{l \text{ times}} \longrightarrow C^{\infty}(M) \qquad \color{blue}{(\star)}
\end{equation}
is induced by a smooth $(k,l)$ tensor field as above iff this map is multilinear over $C^{\infty}(M)$.
In the proof of the above lemma, the tensor induce from the multilinear map $\tau : \mathfrak{X}^*(M) \times \cdots \mathfrak{X}^*(M) \times \mathfrak{X}(M)\times \cdots \mathfrak{X}(M) \longrightarrow C^{\infty}(M)$ is a tensor field $A : M \to T^{(k,l)}TM$ defined as
$$
A_p(w^1,\dots,w^k,v_1,\dots,v_l):= \tau(\omega^1,\dots,\omega^k,X_1,\dots,X_l)(p)
$$
where $\omega^i\in\mathfrak{X}^*(M)$ is any smooth extension of $w^i\in T^*_pM$ and $X_i \in \mathfrak{X}(M)$ is any smooth extension of $v_i \in T_pM$, for each $i$ (the map $A$ independent of the extensions).
Because of this lemma, we often identify $A$ with its induced map $\tilde{A}$.
In your case above the map is
\begin{equation}
R : \mathfrak{X}(M)^* \times \mathfrak{X}(M) \times \mathfrak{X}(M) \times \mathfrak{X}(M) \rightarrow C^{\infty}(M)
\end{equation}
defined as $$R(\omega,X,Y,Z) := \omega(\nabla_X\nabla_Y Z - \nabla_Y\nabla_X Z - \nabla_{[X,Y]}Z)$$ is defines a $(3,1)$-tensor field if you can show that $R$ is linear over $C^{\infty}(M)$ on each arguments. This including the first and the fourth argument which is not mentioned before in the comments above. The first argument is obvious
$$
R(f\omega,\cdot,\cdot,\cdot) = (f\omega)(\cdot) = f\,\omega(\cdot)
$$
The second and the third and the fourth is similar, it follows from
\begin{align*}
\nabla_{fX_1+gX_2}&= f\nabla_{X_1} + g \nabla_{X_2} \quad \textbf{(Linearity)}\\
\nabla_X(fY) &= f\nabla_X Y + (Xf)Y \quad \textbf{(Leibniz Rule)}\\
[fX,Y] &= f[X,Y] -(Yf)X
\end{align*}
For example,
\begin{align*}
R(\omega,fX,Y,Z) &= \omega (\nabla_{fX} \nabla_Y Z - \nabla_Y \nabla_{fX} Z-\nabla_{[fX,Y]}Z) \\
&= \omega (f\nabla_X \nabla_Y Z - \nabla_Y (f\nabla_XZ ) - \nabla_{f[X,Y]-(Yf)X}Z) \\
&=\omega (f\nabla_X \nabla_Y Z - f\nabla_Y\nabla_XZ - \require{cancel}{\cancel{(Yf)\nabla_XZ}} - f\nabla_{[X,Y]} Z + {\cancel{(Yf)\nabla_XZ}}) \\
&= f R(\omega,X,Y,Z).
\end{align*}
This lemma is oftenly used in Riemannian geometry text implicitly. You can see the proof in Lee's book here p.318.
Beside the type of maps in $\color{blue}{(\star)}$ above, another but similar way to spot a disguised tensor field is through these kind of maps
$$
\tau_0 : \underbrace{\mathfrak{X}^*(M)\times\cdots\times\mathfrak{X}^*(M)}_{k \text{ times}} \times \underbrace{\mathfrak{X}(M) \times \cdots \times \mathfrak{X}(M)}_{l \text{ times}} \longrightarrow \color{red}{\mathfrak{X}(M)}.
$$
If this map is multilinear over $C^{\infty}(M)$ then we can define a map
$$
\tau : \underbrace{\mathfrak{X}^*(M)\times\cdots\times\mathfrak{X}^*(M)}_{\color{red}{k+1 \text{ times}}} \times \underbrace{\mathfrak{X}(M) \times \cdots \times \mathfrak{X}(M)}_{l \text{ times}} \longrightarrow C^{\infty}(M)
$$
as $\tau(\omega^{1},\dots,\omega^{k}, \omega^{k+1},X_1,\dots,X_l) = \tau_0(\omega^{1},\dots,\omega^{k},X_1,\dots,X_l)(\omega^{k+1})$, which is also multilinear over $C^{\infty}(M)$. Therefore $\tau_0$ induced by a smooth $(k+1,l)$-tensor field (as its induced map $\tau$ induce them in the Tensor Characterization Lemma above) iff $\tau_0$ is multilinear over $C^{\infty}(M)$. Other respective variants of $\tau_0$ can be deduced the same way.
Best Answer
The abstract definition of the curvature tensor is good as is, see, if $\partial_\mu$ is the $\mu$th coordinate basis vector for some local chart, we have $\nabla_{\partial_\mu}\partial_\nu=\Gamma^\sigma_{\mu\nu}\partial_\sigma$, so we have $$ R(\partial_\mu,\partial_\nu)\partial_\rho=\nabla_{\partial_\mu}\nabla_{\partial_\nu}\partial_\rho-\nabla_{\partial_\nu}\nabla_{\partial_\mu}\partial_\rho=\nabla_{\partial_\mu}(\Gamma^\sigma_{\nu\rho}\partial_\sigma)-\nabla_{\partial_\nu}(\Gamma^\sigma_{\mu\rho}\partial_\sigma)=\partial_\mu\Gamma^\sigma_{\nu\rho}\partial_\sigma+\Gamma^\sigma_{\nu\rho}\nabla_{\partial_\mu}\partial_\sigma-(\mu\leftrightarrow\nu)=\\=\partial_\mu\Gamma^\sigma_{\nu\rho}\partial_\sigma+\Gamma^\sigma_{\nu\rho}\Gamma^\lambda_{\mu\sigma}\partial_\lambda-(\mu\leftrightarrow\nu)=(\partial_\mu\Gamma^\sigma_{\nu\rho}+\Gamma^\sigma_{\mu\lambda}\Gamma^\lambda_{\nu\rho}-(\mu\leftrightarrow\nu))\partial_\sigma, $$ so the formula $$ R(X,Y)Z=\nabla_X\nabla_YZ-\nabla_Y\nabla_XZ-\nabla_{[X,Y]}Z $$ gives the correct curvature formula in the case of torsion as well.
What confuses you is that in index notation, the expression $[\nabla_\mu,\nabla_\nu]$ actually corresponds to an antisymmetric expression made from second covariant derivatives, in abstract language, the torsionless curvature tensor can be expressed as $$ R(X,Y)Z=\nabla^2_{X,Y}Z-\nabla^2_{Y,X}Z, $$ and its THIS formula that needs to be modified for torsion. It's because $\nabla^2_{X,Y}Z$ corresponds to $X^\mu Y^\nu\nabla_\mu\nabla_\nu Z^\sigma$ in index notation. In abstract notation, $\nabla_X\nabla_Y Z\neq\nabla^2_{X,Y}Z$, because in the first term, $\nabla_X$ also acts on $Y$, while for the "second covariant derivatives", $Y$ is applied "from outside the differential operator".
In index notation, $$ \nabla_X\nabla_Y Z\Longleftrightarrow X^\mu\nabla_\mu(Y^\nu\nabla_\nu Z^\sigma) \\ \nabla^2_{X,Y}Z\Longleftrightarrow X^\mu Y^\nu\nabla_\mu\nabla_\nu Z^\sigma,$$ work out for yourself in component notation that these two are not equivalent.
The difference between the antisymmetric form of $\nabla_X\nabla_Y Z$ and $\nabla^2_{X,Y}Z$ will produce a term proportional to $\nabla_{\nabla_YX-\nabla_XY}$, which is $\nabla_{[X,Y]}$ iff $\nabla$ is torsionless, hence the need to modify $R(X,Y)Z=\nabla^2_{X,Y}Z-\nabla^2_{Y,X}Z$ if we have nonvanishing torsion, since the correct curvature tensor is given by $R(X,Y)Z=[\nabla_X,\nabla_Y]Z-\nabla_{[X,Y]}Z$.