To answer your second question first: an orthogonal matrix $O$ satisfies $O^TO=I$, so $\det(O^TO)=(\det O)^2=1$, and hence $\det O = \pm 1$. The determinant of a matrix tells you by what factor the (signed) volume of a parallelipiped is multipled when you apply the matrix to its edges; therefore hitting a volume in $\mathbb{R}^n$ with an orthogonal matrix either leaves the volume unchanged (so it is a rotation) or multiplies it by $-1$ (so it is a reflection).
To answer your first question: the action of a matrix $A$ can be neatly expressed via its singular value decomposition, $A=U\Lambda V^T$, where $U$, $V$ are orthogonal matrices and $\Lambda$ is a matrix with non-negative values along the diagonal (nb. this makes sense even if $A$ is not square!) The values on the diagonal of $\Lambda$ are called the singular values of $A$, and if $A$ is square and symmetric they will be the absolute values of the eigenvalues.
The way to think about this is that the action of $A$ is first to rotate/reflect to a new basis, then scale along the directions of your new (intermediate) basis, before a final rotation/reflection.
With this in mind, notice that $A^T=V\Lambda^T U^T$, so the action of $A^T$ is to perform the inverse of the final rotation, then scale the new shape along the canonical unit directions, and then apply the inverse of the original rotation.
Furthermore, when $A$ is symmetric, $A=A^T\implies V\Lambda^T U^T = U\Lambda V^T \implies U = V $, therefore the action of a symmetric matrix can be regarded as a rotation to a new basis, then scaling in this new basis, and finally rotating back to the first basis.
For the inverse, you can just apply the definition and compute:
$(ab)(b^{-1}a^{-1})=a(bb^{-1})a^{-1}=aa^{-1}=id$
and similarly $(b^{-1}a^{-1})(ab)=id$. Hence $b^{-1}a^{-1}$ is an (and by uniqueness of inverses, the) inverse of $ab$.
A popular way to 'explain' this, is by saying "the inverse of 'putting on socks and then shoes', is 'taking off shoes and then socks'".
For the transpose, you can also apply the definition: if $A$ has matrix elements $(a_{ij})_{ij}$, then $A^T$ has matrix elements $(a_{ji})_{ij}$. Then compute the matrix elements of $(AB)^T$ and of $B^TA^T$ and see that they are equal.
A more conceptual way however is the following: taking the dual is a contravariant functor. That is, a linear map $A:V\to W$ has dual map $A^*:W^*\to V^*$ between the dual spaces (in the other direction!), and this is compatible with composition: $(A\circ B)^*=B^*\circ A^*$. The definition is $A^*=-\circ A$, so just precompose the linear functional with $A$; the composition property is then immediate. Now if $A$ has matrix $a_{ij}$ with repsect to some bases of $V,W$, then the matrix of $A^*$ w.r.t. the corresponding dual bases of $V^*,W^*$ is $a_{ji}$ (this is an easy check by writing out the definitions).
Best Answer
Let me pick the middle two terms $eW'X' + XWe'$. If the two terms result in scalar values then we could take transpose of the two or either one of the terms. so I will take transpose of the first term $(eW'X')' = XWe'$ and this is same as the 2nd term so they could add together to form $2XWe'$