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.
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 $R(X,Y)Z=[\nabla_X,\nabla_Y]Z-\nabla_{[X,Y]}Z$.
Best Answer
What you are asking essentially comes from Gauss's Theorema Egregium.
The theorem states that the Gaussian curvature is given by
$$K=\frac{R_{1212}}{g}$$ where
$$R_{ijkl}=\frac{\partial \Gamma_{ijk}}{\partial x_l}- \frac{\partial \Gamma_{ijl}}{\partial x_k}+ \Gamma^{r}_{jk}\Gamma_{irl}-\Gamma^{r}_{jl}\Gamma_{irk}$$ is the Riemann curvature tensor.
Gauss's theorem can be expressed by many formulas, this is the expression using the Christoffel symbols. Note that $K$ also need $g$, (but only as a denominator), so its not dependent only on the Christoffel symbols alone.
Geometrically the Christoffel symbols arise from the following equations, \begin{align*} \mathbf{r}_{uu}&=\Gamma_{11}^1 \ \mathbf{r}_{u} + \Gamma_{11}^2 \ \mathbf{r}_{v} +L \ \mathbf{N}\\ \mathbf{r}_{uv}&=\Gamma_{12}^1 \ \mathbf{r}_{u} + \Gamma_{12}^2 \ \mathbf{r}_{v} +M \ \mathbf{N}\\ \mathbf{r}_{vv}&=\Gamma_{22}^1 \ \mathbf{r}_{u} + \Gamma_{22}^2 \ \mathbf{r}_{v} +N \ \mathbf{N}\\ \end{align*}
If you examine these equations you see that these coefficients arise from differentiating the tangent vectors, which results in an vector which is no longer tangent, but then projecting the derivative back onto the surface.
What I have just described is exactly covariant differentation. Thus Christoffel symbols and connections are basically the same thing, in fact
$$\nabla_{X_i}(X_i)=\sum_k \Gamma^k_{ji}X_k$$ they are just the coefficients of the connection. Historically the curvature formula came first and latter connections were introduced. In fact it can be proven that the only invariants of a Riemann metric $g_{ij}$ are the Riemann Curvature tensor $R_{ijkl}$ and its covariant derivatives. Thus one way or another the Curvature tensor is forced upon us.
It might be helpful to you to prove the formula in the question in local coordinates using the equations in this answer.
Hope this helps.