I'm reading an article that states that for any finite dimensional manifold $M$ and any $q \in M, w \in T_q M$, there is a canonical isomorphism between $T_{(q, w)}TM$ and $T_q M \times T_q M$ defined by the Levi-Civita connection $\nabla$ on $M$. Of course, they are both vector spaces of the same dimension, so it follows that they are isomorphic. However, I don't see how a canonical isomorphism is constructed through $\nabla$.
In the end, I suspect that this will have something to do with the push-forward of the exponential map. If we suppose for simplicity that $M$ is complete, then consider the map $F: TM \to M \times M$ given by $F(q, w) = (q, \exp_q(w))$. Then $F_\ast: TTM \to T(M\times M)$ is a local diffeomorphism, and so $(F_{\ast})_{(q, w)}: T_{(q, w)} TM \to T_{F(q, w)}(M \times M)$ is a linear isomorphism. From here, we have a canonical isomorphism $T_{F(q, w)}(M \times M) \cong T_q M \times T_{\exp_q(w)}M$, and we can identify $T_{\exp_q(w)}M \cong T_q M$ via parallel transport. Is this correct?
Let $(E,\pi,M)$ be a smooth vector bundle, and fix $q\in M$ and $w_q\in E_q$. If we have a connection on this vector bundle, then we can construct in a "natural" way an isomorphism $T_{w_q}E\cong T_qM\times E_q$. To prove this, recall that a connection on the vector bundle can be equivalently defined as a choice of a subbundle $HE$ which is complementary to the so-called vertical subbundle $VE:= \ker (T\pi:TE\to TM)$.
Just from this alone, we have the (internal) direct sum decomposition $T_{w_q}E=H_{w_q}E\oplus V_{w_q}E$. Hence, we have an isomorphism $T_{w_q}E\cong H_{w_q}E\times V_{w_q}E$. Now, we can easily show from the definitions that $V_{w_q}E= \ker(T\pi_{w_q}:T_{w_q}E\to T_qM)=T_{w_q}(E_q)$, or in words, the vertical space of $E$ at $w_q$ is equal to the tangent space to $E_q$ at the point $w_q\in E_q$. But now recall that $E_q$ is a vector space so it's tangent space is canonically isomorphic to itself. Let us use $\iota_q:V_{w_q}E=T_{w_q}(E_q)\to E_q$ to denote this isomorphism.
Next, I claim that $H_{w_q}E\cong T_qM$. To see this, we consider the mapping $T\pi_{w_q}:T_{w_q}E\to T_qM$. The domain is a direct sum $T_{w_q}E=H_{w_q}E\oplus V_{w_q}E=H_{w_q}E\oplus \ker(T\pi_{w_q})$. Therefore, the restriction $T\pi_{w_q}\bigg|_{H_{w_q}E}:H_{w_q}E\to T_qM$ is injective (we're taking a linear mapping, and restricting it to a subspace complementary to its kernel... i.e we're "ignoring" the kernel, hence injective). But now both these spaces have the same dimension, so the restricted mapping is an isomorphism.
Putting this all together, we have \begin{align} T_{w_q}E= H_{w_q}E\oplus V_{w_q}E\cong H_{w_q}E\times V_{w_q}E\cong T_qM\times E_q. \end{align} Super explicitly, if we let $P_{H,w_q}:T_{w_q}E\to H_{w_q}E,P_{V,w_q}:T_{w_q}E\to V_{w_q}E$ be the projections induced by the direct sum decomposition, then the above isomorphism is \begin{align} T_{w_q}E\to T_qM\times E_q,\qquad \xi\mapsto (T\pi_{w_q}(P_{H,w_q}(\xi)), \iota_q(P_{V,w_q}(\xi))). \end{align} In words, take a vector in $T_{w_q}E$, and decompose (using the connection) into horizontal and vertical components. The horizontal component can be identified with $T_qM$ by applying $T\pi$. The vertical component can be identified with $E_q$ because tangent space to a vector space is the vector space itself.
If you now take the special case of $E=TM$, then you get $T_{w_q}(TM)\cong T_qM\times T_qM$.