Natural transformation between functors $\mathbf {CRing}\to\mathbf{Mon}$ Regarding Example 1.3.5:
1) How to see that a ring homomorphism $R\to S$ induces a monoid homomorphism $M_n(R)\to M_n(S)$? It definitely induces a map, but why is it a monoid homomorphism?
2) Why do the naturality squares commute? Let $R,S$ be rings and let $\phi:R\to S $ be a ring homomorphism. What needs to be checked is that $\det_S\circ M_n(\phi)=U(\phi)\circ \det_R$. Either side is a function on $M_n(R)$. So let $A\in M_nR$; then the LHS evaluated at $A$ is $U(\phi)(\det_R A)=\phi(\det_R A)$. (I suppose $U(\phi)$ is the same as $\phi$ but regarded as a monoid homomorphism.). What's the RHS? Well, $M_n(\phi)(A)=\phi'(A)$ where $\phi'$ is the map induced by $\phi$ from 1) above. I can't see why $\det_S(\phi'(A))=\phi(\det_R A)$.
3) There is a remark that $\det_R$ is a monoid homomorphism. What is that needed for?
 A: $M_n(R)$ is the monoid of all $n\times n$ matrices with entries from $R$ with matrix multiplication as the binary operation. If $f:R\to S$ is a ring homomorphism, then $M_nf$ sends a matrix $A$ in $M_n(R)$ to the matrix $fA$ in $M_n(S).$ That is, $M_nf(A)=f(A),\ $ which is to say $a_{ij}\mapsto f(a_{ij})$. Then, if $B$ is another matrix in $M_n(R),\ M_nf(AB)=fAB=(fA)(fB)=(M_nf(A))(M_nf(B))$, so $M_nf$ is a monoid homomorphism. Thus, $M_n: \text{CRing} \to \text{Mon}$ is a functor with the advertised property. 
Now, $U : \text{CRing} \to \text{Mon}$ is the evident functor that only "remembers" the multiplicative structure on $R\in \text{CRing}$. On arrows $f:R\to S,\ Uf(R)=f(R):r\mapsto f(r).$
For each $R\in \text{CRing},$ there is the determinant function $\text{det}_R : M_n(R) \to  U(R)$, which carries an $n\times n$ matrix to its determinant. Note that $f:R\to S$ is a ring homomorphism, then $f(\det A)=f\left(\sum _{\sigma\in S_n}\text {sgn}\sigma\  a_{1\sigma(1)\cdots }a_{n\sigma(n)}\right)=\left(\sum _{\sigma\in S_n}\text {sgn}\sigma\  f(a_{1\sigma(1)})\cdots f(a_{n\sigma(n)})\right)$ so $f(\det A)=\det f(A).$
Now, suppose $A\in M_n(R).$ We need to check that if $f:R\to S,$ then $Uf(\det_R A)=\det_S(M_n(R)f(A)).$ So we compute 
$Uf(\det_R A)=f(\det_R A)=\det_S f(A)=\det_S(M_n(R)f(A)).$
Finally, to your last question, yes, because $\det_R$ needs to be a proper morphism between objects of of $\text{Mon}.$
