Does precedence matter in constructing parse trees of arithmetic expressions for postfix notation? Table 4 in Problem Solving with Algorithms and Data Structures's
 chapter on Infix, Prefix and Postfix Expressions gives the following examples:

Infix Expression: A + B * C + D 
Prefix Expression: + + A * B C D
Postfix Expression: A B C * + D +


Why is prefix not: 

+ A + * B C D


To show my reason for this question, I'll use lowercase characters combined to represent a completed operation (e.g. + A B becomes ab). Here's my thinking:

+ A + * B C D 
+ A + bc D
+ A bcd
abcd


Here's my thinking applied to the answer supplied by the text:

+ + A * B C D
+ + A bc D
+ abc D
abcd


They both work?

I'm making the same mistake with

Infix Expression: A + B + C + D 
Prefix Expression: + + + A B C D    
Postfix Expression: A B + C + D +


I write postfix as: 

A B C D + + +


...which is wrong (but I don't know why).
What you're missing is the associativity of operators. All operators you mention in your examples are binary. An expression of the form $A + B + C + D$ is interpreted by the parser as $(((A+B)+C)+D)$. When converting such a fully parenthesized expression into prefix form, we apply the rule $(\alpha + \beta) \to +\alpha\beta$. Denote this transformation by $T$. Then
$$
T[(((A+B)+C)+D)] = +T[((A+B)+C)]D = ++T[(A+B)]CD = +++ABCD.
$$
In a completely analogous way, the operator $S$ converting into postfix makes the transformation $S[(\alpha+\beta)] = \alpha\beta+$:
$$
S[(((A+B)+C)+D)] = S[((A+B)+C)]D+ = S[(A+B)]C+D+ = AB+C+D+.
$$

The operator $+$ is associative, but imagine what happens with an operator like $-$. The expression $A-B-C$ is interpreted as $((A-B)-C)$, and its prefix and postfix forms are $--ABC$ and $AB-C-$, respectively. Any other conversion would result in an arithmetically inequivalent expression. That's why associativity matters.

Most operators associate to the left, just like $+$. Exponentiation traditionally associates to the right.

On top of this, we also have operator precedence, which leads to the parsing of $A+B*C+D$ as $((A+(B*C))+D)$. If you apply the transformations $T$ and $S$ you will get the expressions given in the book.