Going off user26857's comment, we provide a counterexample for Proposition 3.14 in the case that $M$ is not finitely generated. Hopefully you can use this to construct a counterexample for Corollary 3.15 as well.
Take $A = \mathbb{Z}$, and let $M$ be the direct sum of $\mathbb{Z}/k\mathbb{Z}$ as $k$ ranges through $\mathbb{N}$. This is not finitely generated as a $\mathbb{Z}$-module (there are infinitely many nonzero summands). Now, what is the annihilator of this module? Well, it is just $0$, since for any integer $n$ different from $0$, $n$ does not act by $0$ on the $\mathbb{Z}/(n+1)\mathbb{Z}$ factor, so $n$ does not act by $0$ on $M$. (If $n$ is negative, look at $\mathbb{Z}/(1-n)\mathbb{Z}$. Strictly, this is not necessary since any ideal of $\mathbb{Z}$ is generated by a positive number, but oh well.)
Now, let us localize at $S = \mathbb{Z} \setminus \{0\}$, so $S^{-1}A = \mathbb{Q}$ and $S^{-1}\text{Ann}(M) = S^{-1}0 = 0$.
But what is $S^{-1}M$? We claim that it is $0$. This is not hard to show. Let $e_k$ be the generator of the copy of $\mathbb{Z}/k\mathbb{Z}$ in $M$. Then the set $\{e_k\}$ is a $\mathbb{Z}$-generating set for $M$ in the sense that every element of $M$ is a finite linear combination of the elements $e_k$ with coefficients in $\mathbb{Z}$, so $\{e_k/1\}$ is a $\mathbb{Q}$-generating set for $S^{-1}M$ (in this same sense).
But for each $k$, we have $k \cdot e_k = 0$ in $M$, which shows that $e_k/1 = 0$ in $S^{-1}M$. To be a little more explicit, we want to show that $e_k/1 = 0/1$ in $S^{-1}M$. By definition, this happens if and only if there is some $s \in S$ such that $s(1 \cdot e_k - 0 \cdot 1) = 0$. Now take $s = k$.
So now, we have that $\{0\}$ is a $\mathbb{Q}$-generating set for $S^{-1}M$. Thus, $S^{-1}M = 0$, so $\text{Ann}(S^{-1}M) = \mathbb{Q}$, which is different from $S^{-1}\text{Ann}(M) = 0$. (Here, we need the crucial fact that there exist nonzero rational numbers, i.e., that $\mathbb{Q}$ is not equal to $0$.)
Note that your $P_1, \dotsc, P_k$ are the minimal primes of $R$. Note that they are only finite if $R$ is noetherian, you have to add this assumption as the other answer points out.
For $a)$ you should show the following:
If we localize $\prod_{i=1}^k K(R/P_i)$ at $U$, then the component
$K(R/P_i)$ survives if and only if $U \cap P_i = \emptyset$.
Using this, we get:
$K(R)[U^{-1}]$ is the product of all $K(R/P_i)$, where $U \cap P_i=\emptyset$.
Now we start with the reduced ring $R[U^{-1}]$. The minimal primes of this ring are all $P_i[U^{-1}]$ with $P_i \cap U = \emptyset$, i.e. its total quotient ring is the product of all $K(R[U^{-1}]/P_i[U^{-1}])$, where $U \cap P_i=\emptyset$.
Localization commutes with quotients, hence we have
$$K(R[U^{-1}]/P_i[U^{-1}])=K((R/P_i)[U^{-1}])=K(R/P_i),$$
which gives you the result.
The latter equality holds, since we have $(R/P_i)[U^{-1}]=R/P_i$ if $U \cap P_i=\emptyset$.
Best Answer
First, the universal property says that for any homomorphism $f: A\to B$ such that $f(S)\subset B^*$, the grup of units of $B$, there is a unique homomorphism $f': A[S^{-1}]\to B$ such that $f=f'\circ j_S$.
Now, on one hand we have a homomorphism $jj: A[S^{-1}]\to A[S^{-1}][j_S(T)^{-1}]$ and the composition $h=jj\circ j_S$ send clearly the elements of $T$ to invertible elements. So by the universal property, there exist a unique homomorphism say $h': A[T^{-1}] \to A[S^{-1}][j_S(T)^{-1}]$ such that $h=h'\circ j_T$.
On the other hand, the homomorphism $j_T$ sends the elements of $S$ to invertible elements, so we have a map $j'_T: A[S^{-1}]\to A[T^{-1}]$ such that $j_T=j_T'\circ j_S$. This homomorphism sends the elements of $j_S(T)$ to invertible elements, so we have another homomorphism $ i: A[S^{-1}][j_S(T)^{-1}] \to A[T^{-1}]$ such that $j_T'=i\circ jj$.
Now, one only needs to show that $i\circ h'$ is the identity, as well as $h'\circ i$. But this is deduced from the unicity of the universal property.
For the first composition: from $h=jj\circ j_S$, $h=h'\circ j_T$, $j_T'=i\circ jj$ and $j_T=j_T'\circ j_S$, we have $$j_T=(i\circ jj) \circ j_S=i\circ (jj \circ j_S)=i\circ h=(i\circ h')\circ j_T.$$ So $i\circ h':A[T^{-1}]\to A[T^{-1}]$ is a map that verifies the property of $(j_T)'$ corresponding to $j_T$ (it is different from $j'_T$, sory for the messing notation), which also the map $\operatorname{id}$ verifies. So by unicity it must be the identity.
For the second composition $$h'\circ i: A[S^{-1}][j_S(T)^{-1}] \to A[S^{-1}][j_S(T)^{-1}] $$ note first that $jj=h'\circ j'_T$, by its universal property, since $$h'\circ j'_T\circ j_S=h'\circ j_T=h=jj\circ j_S$$ and hence it verifies the same property that $jj$, which is uniquely determined by this property. But then we have $$jj=h'\circ j'_T=h'\circ (i\circ jj)=(h'\circ i) \circ jj$$ and we can argue as the first composition.