Let $(M,g)$ be a (pseudo-)Riemannian manifold. I am having some struggles to relate two different approaches, one based on the abstract index notation, and the other one based on global, coordinate-free, notation.
Consider a general covariant $k$-tensor field $T\in\Gamma^{\infty}(T^{\ast}M^{\otimes k})$. In coordinates, I can write down the components of $T$ as $T_{\mu_{1}\dots\mu_{k}}$. Then, the (connection) Laplacian of $T$ is the $k$-tensor field $\square T\in\Gamma^{\infty}(T^{\ast}M^{\otimes k})$ with components
$$(\square T)_{\mu_{1}\dots\mu_{k}}:=g^{\mu\nu}\nabla_{\mu}\nabla_{\nu}T_{\mu_{1}\dots\mu_{k}}$$
I won't write down the explicit expression of the right-hand side, but I think it is clear what it means. For example, if $k=1$, then
$$(\square T)_{\alpha}=g^{\mu\nu}(\partial_{\mu}\nabla_{\nu}T_{\alpha}-\Gamma^{\lambda}_{\mu\nu}\nabla_{\lambda}T_{\alpha}-\Gamma^{\lambda}_{\mu\alpha}\nabla_{\nu}T_{\lambda})$$
and substituting the usual expression $\nabla_{\mu}T_{\alpha}=\partial_{\mu}T_{\alpha}-\Gamma^{\lambda}_{\mu\alpha}T_{\lambda}$, which gives a lengthy formula for the components of the Laplacian.
On the other hand, on wikipedia for example, the connection Laplacian is defined globally as
$$\square T:=\mathrm{tr}(\nabla^{2}T)$$
with the double covariant derivative $\nabla_{X,Y}^{2}T:=\nabla_{X}\nabla_{Y}T-\nabla_{\nabla_{X}Y}T$. I am a bit confused about the term $\nabla_{\nabla_{X}Y}T$. The first term $\nabla_{X}\nabla_{Y}T$, using the trace, should give my coordinate expression above, i.e. $g^{\mu\nu}\nabla_{\mu}\nabla_{\nu}T_{\alpha}$, but what about the second term? Or does it vanish?
Maybe my confusion is related to how to relate the expression $\nabla_{\alpha}\nabla_{\beta}T_{\mu_{1}\dots\mu_{k}}$ to the double covariant derivative $\nabla^{2}T$...
By definition a connection (as defined here) is a $\mathbb{R}$ linear map $$\nabla: \Gamma ( T^{\ast}M^{\otimes k}) \to \Gamma (T^{\ast}M \otimes T^{\ast}M^{\otimes k})$$ so that $$ \nabla (fs) = df \otimes s + f \nabla s $$ for all $f \in C^\infty (M)$ and $s \in \Gamma(T^{\ast}M^{\otimes k})$.
In our case a connection is naturally defined for any tensor bundle (the levi civita connection) and i will denote all of them with $\nabla$. Now the second covariant derivative is simply $\nabla $ applied twice and the laplacian is the (metric) contraction of the first two factors of the second covariant derivative (this is the definition in the linked wikipedia article).
The definition "$\square T:=\mathrm{tr}(\nabla^{2}T)$" is sloppy, firstly it is unclear which factors are supposed to be contracted and furthermore we need to change the type of one of the factors in the tensor to be able to take the trace (we need a covariant and a contravariant factor to contract).
In abstract index notation: Let $t_{bcd \dots } \in T^{\ast}M^{\otimes k}$. We write $\nabla_a t_{ b c \dots}$ for the resulting tensor when the connection is applied to $t_{bcd \dots }$, which makes sense because $\nabla_a t_{ b c \dots}$ has an additional covariant factor.
So then (i use $\Delta$ for the laplacian) $$\Delta t_{bcde \dots } = \nabla_a \nabla^a t_{bcde \dots }.$$ When computing the components we arrive at the same expression as you have for the components of $\Box T$.
To the second question regarding "$\nabla^2_{X,Y}T$":
Now for any $p \in M$ there is an isomorphism $J$: $$ \begin{align*} J: T_p^\ast M \otimes T_p^\ast M^{\otimes k} &\longmapsto L(T_pM ,T_p^\ast M^{\otimes k}) \\ t_{a b c \dots} &\longmapsto \big (v^a \mapsto v^a t_{abc \dots} \big) \end{align*} $$ That is to say that we simply contract the first slot (the same works with more factors of $T_p^\ast M$).
Now (by definition) $(\nabla^2_{X,Y}T)_p = J(T_p)(X_p,Y_p)$. In abstract index notation (on the left) this quantity is (suppressing $p$ as usual) $$\nabla^2_{v,w}t = v^a w^b \nabla_a \nabla_b t_{cde \dots}$$ Now check the fourth equation here and we can see that this is the same as the definition for $\nabla^2_{X,Y}T$ you cited.