vela said:
If you combine a finite number of infinitesimals, you still get an infinitesimal. To get a finite rotation, you have to combine an infinite number of infinitesimal rotations. It's the transition from finite to infinite where you lose the commutativity.
You may just need to rely on the math here. I don't think you're going to find a simple or intuitive explanation. Mathematicians will tell you that just because a statement is true for finite cases, you can't assume it's true for the infinite case. There are other examples that I can't recall at the moment, but I don't think it's unusual.
No, it's very simple!
Again: rotations around a fixed axis commute, no matter whether they are finite or you use an infinitesimal approximation.
As an alternative to my previous derivation in #13 of the matrix-exponential which maps the Lie algebra to the Lie group you may consider a rotation by a finite angle ##\varphi## around an axis given by the unit vector ##\vec{n}## (the usual right-hand rule applies).
Make ##N## a large natural number. Then ##\varphi/N## can be made arbitrarily small and you can use the linear approximation
$$\hat{R}(\vec{n},\varphi/N)=(\hat{1} + \vec{n} \cdot \hat{\vec{t}} \varphi/N).$$
Now
$$\hat{R}(\vec{n},\varphi_1+\varphi_2)= \hat{R}(\vec{n},\varphi_1) \hat{R}(\vec{n},\varphi_2),$$
which also shows formally that the rotations around one fixed axis commute. This equation can be used to write
$$\hat{R}(\vec{n},\varphi)=\hat{R}(\vec{n},N \varphi/N) = [\hat{R}(\vec{n},\varphi/N)]^N.$$
In the limit ##N \rightarrow \infty## the error made when using the linear approximation for ##\hat{R}(\vec{n},\varphi/N)## becomes arbitrarily small and thus you can write
$$\hat{R}(\vec{n},\varphi) = \lim_{N \rightarrow \infty} \left (\hat{1} + \vec{n} \cdot \hat{\vec{t}} \frac{\varphi}{N} \right) ^{N}=\exp(\vec{n} \cdot \hat{\vec{t}} \varphi).$$