Concatenation is not commutative: if $\alpha$ and $\beta$ are finite strings of symbols, it is generally not true that $\alpha\beta=\beta\alpha$. For example, $a^*b$ represents the set of words $\{b,ab,aab,aaab,\ldots\}$, and $ba^*$ represents the set of words $\{b,ba,baa,baaa,\ldots\}$; clearly these are not the same set, so $a^*b\ne ba^*$. Your next three examples are similar: $a^mb^n$ is a string of $m$ $a$s followed by a string of $n$ $b$s, while $b^na^m$ is exactly the opposite, a string of $n$ $b$s followed by a string of $m$ $a$s. For instance, if $m=2$ and $n=3$ these are $aabbb$ and $bbbaa$, which are certainly not the same string. You should try to see why $a^mb^n=b^na^m$ if and only if at least one of $m$ and $n$ is $0$.
Union is commutative, so it’s quite true that the regular expressions $a+b$ and $b+a$ are equivalent; concatenation is not commutative, so the regular expressions $ab$ and $ba$ are not equivalent. If you’ve been that $ab=ba$, you’ve been taught incorrectly.