A composition is, in essence, applying one function after another (to form a "chain of functions"). From a symbolic/algebraic perspective, you are "plugging one function into the other".
In precalculus-related courses, when we write f(g(x))f(g(x)), the notation means to first apply the function gg to the number xx to get the number g(x)g(x), then apply the function ff to the number g(x)g(x) to get f(g(x))f(g(x)) (so we work "from inside out"). (In some other math courses, the inputs and outputs don't have to be numbers).
For the function k(x)=e^(sin(x))k(x)=esin(x), when a number xx is substituted, the sine function must be evaluated first to get the number sin(x)sin(x). This number then must be made the power of ee to get the final output e^(sin(x))esin(x). Therefore, the sine function must be the "inner" (first applied) function and the exponential function must be the "outer" (second applied) function.
Hence, k(x)=f(g(x))k(x)=f(g(x)) when f(x)=e^(x)f(x)=ex and g(x)=sin(x)g(x)=sin(x).
Extra Info (ignore this at the moment if it's confusing): An alternative notation for this is to use the "circle operator" and write k(x)=(f circ g)(x)k(x)=(f∘g)(x). This can also be written as k=f circ gk=f∘g, which emphasizes that circ∘ is a "binary operator" that takes two given functions ff and gg and creates a new function f circ gf∘g. In general, the circle operator circ∘ is not commutative (that is, in general, f circ g != g circ ff∘g≠g∘f. For the present example, these are indeed not equal because e^(sin(x)) !=sin(e^(x))esin(x)≠sin(ex) as functions (though their graphs do cross each other). Order of operations matters!