Question 1 The two procedures indeed give the same Lie algebroid. One possible way of seeing this is by considering the flows of vector fields: a section of $T(GL(E))/GL(n)$ is a vector field on the frame bundle that is $GL(n)$-invariant, and this means that its flow is by $GL(n)$-equivariant diffeomorphisms. But this is, in turn, equivalent to a family of vector bundle isomorphisms of $E$. When we differentiate this family, we obtain a linear vector field on $E$.

More conceptually, the categories of $GL(n)$-principal bundles and rank $n$-vector bundles (where we only consider automorphisms) are equivalent. Therefore the symmetries are the same, and in particular, the infinitesimal symmetries are the same. But these are precisely the $GL(n)$-invariant vector fields on the one hand, and linear vector fields on the other.

The link to differential operators is also not so difficult to see: a section of $E^{*}$ is the same as a linear function on $E$. So given a section of $E^*$, we may differentiate it by a vector field on $E$. In general this will not return a section of $E^*$, rather it will just be some function on $E$. Linear vector fields are precisely the vector fields on $E$ that send linear functions to linear functions. So we can think of linear vector fields as differential operators for $E^*$ (and by dualizing, for $E$ as well).

Question 2 I believe the reason Atiyah originally introduced the Atiyah algebroid was to study the question of the existence of holomorphic connections. Note that connections $\nabla$ are a type of differential operator: given a vector field $X$, then we get a first-order differential operator $\nabla_{X} : \Gamma(E) \to \Gamma(E)$, whose symbol is the vector field $X$.

The Atiyah algebroid sits in a nice short exact sequence of Lie algebroids:

$0 \to End(E) \to At(E) \to TM \to 0. $

The right-hand map is the anchor. From the perspective of principal bundles, this is the map you constructed in your question, and the kernel is the 'vertical bundle'. From the perspective of differential operators, where we view sections of the Atiyah algebroid as differential operators on $E$ of order $1$, the anchor map becomes the symbol. Hence, the kernel $End(E)$, corresponds to the differential operators of order $0$.

Looking back at the definition of a connection, you can see that it is precisely giving you a section of the symbol (i.e. anchor) map. So the existence of a connection on $E$ is exactly the same thing as an isomorphism $At(E) \cong End(E) \oplus TM$ (respecting the extension structure). In the world of smooth manifolds, short exact sequences of bundles always split: every bundle admits a connection. In the holomorphic world this is no longer the case: the existence of holomorphic connections is measured by the extension class of $At(E)$, which is a cohomology class in $H^{1}(T^{*}M \otimes End(E))$, called the Atiyah class of $E$.

Question 1. The two procedures indeed give the same Lie algebroid. One possible way of seeing this is by considering the flows of vector fields: a section of $T(GL(E))/GL(n)$ is a vector field on the frame bundle that is $GL(n)$-invariant, and this means that its flow is by $GL(n)$-equivariant diffeomorphisms. But this is, in turn, equivalent to a family of vector bundle isomorphisms of $E$. When we differentiate this family, we obtain a linear vector field on $E$.

More conceptually, the categories of $GL(n)$-principal bundles and rank $n$-vector bundles (where we only consider automorphisms) are equivalent. Therefore the symmetries are the same, and in particular, the infinitesimal symmetries are the same. But these are precisely the $GL(n)$-invariant vector fields on the one hand, and linear vector fields on the other.

The link to differential operators is also not so difficult to see: a section of $E^{*}$ is the same as a linear function on $E$. So given a section of $E^*$, we may differentiate it by a vector field on $E$. In general this will not return a section of $E^*$, rather it will just be some function on $E$. Linear vector fields are precisely the vector fields on $E$ that send linear functions to linear functions. So we can think of linear vector fields as differential operators for $E^*$ (and by dualizing, for $E$ as well).

Question 2. The question of whether principal bundles or differential operators are 'easier' is probably a matter of one's background. But the Atiyah algebroid is useful for studying both.

I believe the reason Atiyah originally introduced the Atiyah algebroid was to study the question of the existence of holomorphic connections. Note that connections $\nabla$ are a type of differential operator: given a vector field $X$, then we get a first-order differential operator $\nabla_{X} : \Gamma(E) \to \Gamma(E)$, whose symbol is the vector field $X$.

The Atiyah algebroid sits in a nice short exact sequence of Lie algebroids:

$0 \to End(E) \to At(E) \to TM \to 0. $

The right-hand map is the anchor. From the perspective of principal bundles, this is the map you constructed in your question, and the kernel is the 'vertical bundle'. From the perspective of differential operators, where we view sections of the Atiyah algebroid as differential operators on $E$ of order $1$, the anchor map becomes the symbol. Hence, the kernel $End(E)$, corresponds to the differential operators of order $0$.

Looking back at the definition of a connection, you can see that it is precisely giving you a section of the symbol (i.e. anchor) map. So the existence of a connection on $E$ is exactly the same thing as an isomorphism $At(E) \cong End(E) \oplus TM$ (respecting the extension structure). In the world of smooth manifolds, short exact sequences of bundles always split: every bundle admits a connection. In the holomorphic world this is no longer the case: the existence of holomorphic connections is measured by the extension class of $At(E)$, which is a cohomology class in $H^{1}(T^{*}M \otimes End(E))$, called the Atiyah class of $E$.