Hint. Try to understand why the Weingarten map $L: T_{p}{\mathbb{S}^2} \rightarrow T_{p}{\mathbb{S}^2}$ on the sphere is given by $$L=-\frac{1}{r}\operatorname{id}$$ (I assume that $r$ is the radius of your sphere), and then use the Gauss equation
$$
R(X,Y,Z,W) = <II(Y,Z),II(X,W)> - <II(X,Z),II(Y,W)>
$$
where $II(X,Y) = <L(X),Y>$ is the second fundamental form.
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
This seems like mainly a question about what the abstract index notation is asking you to do. Let's look at a single term from the right hand: $\nabla_X\nabla_YZ$. Recall that in abstract index notation, two tensors juxtaposed signifies their tensor product. For example, if $V^a$ and $W^b$ are vector fields, $V^aW^b$ is supposed to mean the tensor field $V \otimes W$. Pairing of an upper and lower index signifies a trace: if $\varphi_a$ is a 1-form and $V^b$ is a vector field, $\varphi_aV^a =\operatorname{tr}(\varphi\otimes V) = \varphi(V)$. So the way we're supposed to think of this is first taking their tensor product and then tracing. Lastly, notice that $\nabla_VW = \operatorname{tr}(\nabla V \otimes W)$ when considered as a section of $T^* \otimes T \otimes T$, where we trace over the first and last slots.
Returning to the first term above, we then have:
\begin{align} \nabla_X\nabla_YZ & = X^c\nabla_cY^d\nabla_dZ^b \\ & = \nabla_X(\operatorname{tr}(Y \otimes \nabla_-Z)) \\ & = \operatorname{tr}(\nabla_XY \otimes \nabla_-Z + Y \otimes \nabla_X\nabla_-Z)) \\ & = Y^dX^c \nabla_c\nabla_dZ^b + (X^c\nabla_cY)^d\nabla_dZ^b \\ & = \nabla^2_{X,Y}Z + \nabla_{\nabla_XY}Z. \end{align}
And from here the rest shouldn't be bad.