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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.

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.