Why can we ignore the constant factor in Weis's proof of the Master Theorem
Suppose that your function is always in the range $cN^k$ to $CN^k$. Consider the three recurrences
$$
s(N) = as(N/b) + cN^k, S(N) = aS(N/b) + CN^k, R(N) = aR(N/b) + N^k.
$$
For appropriate initial conditions, $s(N) \leq T(N) \leq S(N)$ will hold, as a straightforward proof by induction would show. On the other hand, for appropriate initial conditions, $s(N) = cR(N)$ and $S(N) = CR(N)$, as another straightforward induction shows. This explains Weis's remark.

What it happening here? As you unroll the recursion, you add various terms which are $\Theta(N^k)$, for various values of $N$. The contribution of the big Theta to the final sum is at most a constant factor. Instead of $a^m \sum_{i=0}^m (\frac{b^k}{a})^i$, you get $\Theta\bigl(a^m \sum_{i=0}^m (\frac{b^k}{a})^i\bigr)$. Crucially, the additional constant factor is global and doesn't accumulate throughout the recursion.

As an example, consider the generic recurrence $s(N) = 2s(N-1) + f(N)$, with base case $s(0) = 0$. The solution is $$ s(N) = f(N) + 2f(N-1) + 4f(N-2) + \cdots + 2^{N-1} f(1). $$ If we multiply $f(N)$ by a constant factor, the solution of this recurrence will be multiplied by the same constant factor, while the constant $2$ stays the same.