Let's say that I'm viewing a 10-meter-long object from 2 kilometers away using a pair of binoculars with 160x magnification.
This calculator states that, to the naked eye, a 10-meter object at 2 kilometers has an apparent size of 0.286 degrees.
From my admittedly not-entirely-watertight understanding of apparent size and arcs, the effect of the binoculars can be represented by the equation 0.286 * 160 = 45.76, meaning that that 10-meter object would instead have an apparent size of 45.76 degrees.
However, I feel like I'm missing something here. Am I?
The calculator is mistaken: the $0.286^\circ$ that it returns is the actual field of view, not the apparent field of view. (By elementary trigonometry, this angle is just $2\arctan\frac{10/2}{2000}.$)
Your equation is incorrect: $$\text{apparent field of view}\neq\text{magnification $\times$ actual field of view}.$$ Instead, by elementary right-angled trigonometry, $$\text{magnification}\\=\frac{\text{actual distance}}{\text{apparent distance}}\\=\left(\frac{\text{object height}}2\div\tan\frac{\text{apparent field of view}}2\right)\\\div\left(\frac{\text{object height}}2\div\tan\frac{\text{actual field of view}}2\right),$$ so $$\text{apparent field of view}\\=2\arctan\left(\text{magnification}\times\tan\frac{\text{actual field of view}}2\right)\\=43.6^\circ.$$