This doubt stems from the fact that divergence is a dot product.
Take some vector field $\vec{\mathbf{F}}(x, y) = u(x, y)\mathbf{i} + v(x, y)\mathbf{j}$, where $u$ and $v$ are ordinary scalar functions (OSFs).
Now, we know that $\nabla = \frac{\partial}{\partial x}\mathbf{i} + \frac{\partial}{\partial y}\mathbf{j} $
We also know that $(a\mathbf{i} + b\mathbf{j}) \cdot (f\mathbf{i} + g\mathbf{j}) = af + bg$
Now, the divergence of $\vec{\mathbf{F}}$, which is $\text{div}\vec{\mathbf{F}}$, will be the nabla dotted with the vector field:
$$ \nabla \cdot \vec{\mathbf{F}} = \frac{\partial}{\partial x}u + \frac{\partial}{\partial y}v$$
Now, I know that this is just supposed to be differentiating each part with respect to each variable (the thing I memorized), but I also know we are supposed to be multiplying the components.
It seems weird to me as the notions of multiplication and applying operators had never coincided for me before.
Question Is multiplying an operator by a function the same as applying the operator to the function?
Help is appreciated!
Confusion of this sort can only arise when you don't define precisely the terms you're using. In your case, the notion of "multiplication" is not defined, so what is meant by it? Because you can define various notions of multiplication - for instance multiplication of two real numbers is defined differently than multiplication of two complex numbers. Also in your example, I hope you realize that the divergence is $\nabla \cdot$ (i.e. nabla WITH the dot product) and when it is applied to a vector field it is the divergence of that vector field $\nabla \cdot \vec{F}$. The nabla operator on its own (without the dot product) is another operator, which is the gradient operator, since when it's applied to a function it is $\nabla f$ which we know as the gradient of a function. My point is that no notion of multiplication is needed here.