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Yes, OP is right: we understand composition of operators $\hat{A}$ and $\hat{B}$ as $$(\hat{A}\circ \hat{B})(v)~:=~ \hat{A}(\hat{B}(v)), \tag{1}$$ where $v$ is a vector. Note that the composition symbol "$\circ$" is often not written explicitly.

Notabene: Note that if $|x\rangle$ denotes the position ket with eigenvalue $x$, $$ \hat{x}|x\rangle ~=~x |x\rangle, \tag{2}$$ then the convention (1) implies that $$ \hat{p}\hat{x}|x\rangle ~~\stackrel{(1)+(2)}{=}~x \hat{p}|x\rangle \tag{3},$$ because $\hat{p}$ is a linear operator. Together with the CCR $$ [\hat{x},\hat{p}]~=~i\hbar{\bf 1}\tag{4}$$ this implies that $$ \hat{x}\hat{p}|x\rangle ~\stackrel{(3)+(4)}{=}~x \hat{p}|x\rangle +i\hbar|x\rangle \tag{5}.$$ See also this Phys.SE post.

Yes, OP is right: We understand composition of operators $\hat{A}$ and $\hat{B}$ as $$(\hat{A}\circ \hat{B})(v)~:=~ \hat{A}(\hat{B}(v)), \tag{A}$$ where $v$ is a vector. Note that the composition symbol "$\circ$" and parenthesis "$()$" are often not written explicitly. This agrees with OP's eqs. (1) & (2). So far so good.

Things get surprisingly intricate, when we implement this rule (A) on the Dirac notation. Then we define $$\psi(x)~:=~\langle x | \psi \rangle ~=~| x \rangle^{\dagger}| \psi \rangle, \tag{B}$$ $$ (\hat{A} (\psi))(x)~:=~ \langle x |\hat{A}| \psi \rangle ~=~\left(\hat{A}^{\dagger}| x \rangle\right)^{\dagger}| \psi \rangle, \tag{C} $$ $$ ((\hat{A}\circ \hat{B})(\psi))(x) ~:=~ \langle x |(\hat{A}\circ \hat{B})| \psi \rangle ~=~\left((\hat{B}^{\dagger}\circ \hat{A}^{\dagger})| x \rangle\right)^{\dagger}| \psi \rangle, \tag{D}$$ and so forth. Here $|x\rangle$ denotes the position ket state with eigenvalue $x$, $$ \hat{x}|x\rangle ~=~x |x\rangle. \tag{E}$$

Let us for simplicity assume that the operators $\hat{A}$, $\hat{B}$, etc, are self-adjoint. If one has never seen the lhs. of eq. (D) before, one might worry that the operators $\hat{A}$ and $\hat{B}$ seem to be composed in the wrong order! It turns out that in the end, it works out correctly after all. See e.g. the next example.

Example: The convention (A) implies that $$ \hat{p}\circ \hat{x}|x\rangle ~\stackrel{(A)+(E)}{=}~x \hat{p}|x\rangle \tag{F},$$ because $\hat{p}$ is a linear operator. Therefore, we calculate $$ ((\hat{x}\circ \hat{p})(\psi))(x) ~\stackrel{(D)}{=}~\left((\hat{p}\circ \hat{x})| x \rangle\right)^{\dagger}| \psi \rangle ~\stackrel{(F)}{=}~x\left(\hat{p}| x \rangle\right)^{\dagger}| \psi \rangle ~\stackrel{(C)}{=}~x(\hat{p}( \psi))(x),$$ as it should be. See also my related Phys.SE answer here.

Yes, OP is right: we understand composition of operators $\hat{A}$ and $\hat{B}$ as $$(\hat{A}\circ \hat{B})(v)~:=~ \hat{A}(\hat{B}(v)), \tag{1}$$ where $v$ is a vector. We often don't write the composition symbol "$\circ$" explicitly.

Notabene: Note that if $|x\rangle$ denotes the position ket with eigenvalue $x$, $$ \hat{x}|x\rangle ~=~x |x\rangle, \tag{2}$$ then the convention (1) implies that $$ \hat{p}\hat{x}|x\rangle ~~\stackrel{(1)+(2)}{=}~x \hat{p}|x\rangle \tag{3},$$ because $\hat{p}$ is a linear operator. Together with the CCR $$ [\hat{x},\hat{p}]~=~i\hbar{\bf 1}\tag{4}$$ this implies that $$ \hat{x}\hat{p}|x\rangle ~\stackrel{(3)+(4)}{=}~x \hat{p}|x\rangle +i\hbar|x\rangle \tag{5}.$$ See also this Phys.SE post.

Yes, OP is right: we understand composition of operators $\hat{A}$ and $\hat{B}$ as $$(\hat{A}\circ \hat{B})(v)~:=~ \hat{A}(\hat{B}(v)), \tag{1}$$ where $v$ is a vector. Note that the composition symbol "$\circ$" is often not written explicitly.

Notabene: Note that if $|x\rangle$ denotes the position ket with eigenvalue $x$, $$ \hat{x}|x\rangle ~=~x |x\rangle, \tag{2}$$ then the convention (1) implies that $$ \hat{p}\hat{x}|x\rangle ~~\stackrel{(1)+(2)}{=}~x \hat{p}|x\rangle \tag{3},$$ because $\hat{p}$ is a linear operator. Together with the CCR $$ [\hat{x},\hat{p}]~=~i\hbar{\bf 1}\tag{4}$$ this implies that $$ \hat{x}\hat{p}|x\rangle ~\stackrel{(3)+(4)}{=}~x \hat{p}|x\rangle +i\hbar|x\rangle \tag{5}.$$ See also this Phys.SE post.

Yes, OP is right: we understand composition of operators $\hat{A}$ and $\hat{B}$ as $$(\hat{A}\circ \hat{B})(v)~:=~ \hat{A}(\hat{B}(v)), \tag{1}$$ where $v$ is a vector.

Notabene: Note that if $|x\rangle$ denotes the position ket with eigenvalue $x$, $$ \hat{x}|x\rangle ~=~x |x\rangle, \tag{2}$$ then the convention (1) implies that $$ \hat{p}\hat{x}|x\rangle ~=~x \hat{p}|x\rangle \tag{3},$$ because $\hat{p}$ is a linear operator. See also this Phys.SE post.

Yes, OP is right: we understand composition of operators $\hat{A}$ and $\hat{B}$ as $$(\hat{A}\circ \hat{B})(v)~:=~ \hat{A}(\hat{B}(v)), \tag{1}$$ where $v$ is a vector. We often don't write the composition symbol "$\circ$" explicitly.

Notabene: Note that if $|x\rangle$ denotes the position ket with eigenvalue $x$, $$ \hat{x}|x\rangle ~=~x |x\rangle, \tag{2}$$ then the convention (1) implies that $$ \hat{p}\hat{x}|x\rangle ~~\stackrel{(1)+(2)}{=}~x \hat{p}|x\rangle \tag{3},$$ because $\hat{p}$ is a linear operator. Together with the CCR $$ [\hat{x},\hat{p}]~=~i\hbar{\bf 1}\tag{4}$$ this implies that $$ \hat{x}\hat{p}|x\rangle ~\stackrel{(3)+(4)}{=}~x \hat{p}|x\rangle +i\hbar|x\rangle \tag{5}.$$ See also this Phys.SE post.