Joshua gave nice argument, let me provide my perspective on 1, with several arguments. The bottom line is that the constant $k^{\ell}$ comes from $f(1)=k^{\ell}$, and that very little information on $f$ is needed in order to derive this bound.

**Claim 1:** Let $F(x) = \sum_{\nu} a_{\nu} x^{\nu} = 1+a_{1}x+\ldots$ be a power series with radius of convergence $r>1/2$. Then
$$F(1/p) \le \left(1+\frac{1}{p}\right)^{a_1} \left(1+\frac{1}{p^2}\right)^{C}$$
for large enough $C$ depending on $F$.

**Proof:** Consider $$G(x) = \frac{F(x)(1+x)^{-a_1}-1}{x^2}.$$ It is analytic in $|x|<\min\{r,1\}$. Letting $C= \max\{1 , \max_{[0,1/2]}|G(x)|\}$ we find that
$$F(x) = (1+x)^{a_1}(x^2 G(x)+1) \le (1+x)^{a_1}(1+Cx^2) \le (1+x)^{a_1}(1+x^2)^C$$
for $x \in [0,1/2]$. We now specialize to $x=1/p$. $\blacksquare$

Just apply the claim with $a_{\nu} = f(p^{\nu})$. In your case, however, there is much more structure. For instance, $\sigma_p(f)$ is a rational function of $1/p$, due to the following.

**Claim 2:** For a non-zero polynomial $P$ we have that $(\sum_n P(n) x^n )(1-x)^{\deg P +1}$ is a polynomial in $x$.

**Proof:** By linearity it suffices to consider $P_d(t)=\binom{t+d}{t}$ in which case $(\sum_n P_d(n) x^n) (1-x)^{d+1} \equiv 1$ by the binomial series.

This does not directly help us find an explicit value of $C$. Here is a strategy to get hold on such $C$, in the spirit of Joshua's argument.

**Claim 3:** $\binom{n+k-1}{n}^{\ell} \le \binom{n+k^{\ell}-1}{n}$.
**Proof I:** Since $\binom{a+b}{a} = \prod_{i=1}^{a} \left(1+\frac{b}{i}\right)$, the inequality can be written as $\prod_{i=1}^{n} \left(1+\frac{k-1}{i}\right)^{\ell} \le \prod_{i=1}^{n} \left(1+\frac{k^{\ell}-1}{i}\right)$. It suffices to show $\left(1+\frac{k-1}{i}\right)^{\ell} \le 1+\frac{k^{\ell}-1}{i}$ for each $i$. This follows from Hölder: if $X$ is a random variable assuming the values $1$ with probability $1-1/i$ and $k$ with probability $1/i$, the inequality just says $(\mathbb{E}X)^{\ell} \le \mathbb{E}(X^{\ell})$. **Proof II:** We can deduce the claim from a repeated application of the inequality $\binom{n+k_1-1}{n}\binom{n+k_2-1}{n}\le \binom{n+k_1 k_2-1}{n}$. This has a combinatorial interpretation (see Max's comment to this answer).

**Claim 4:** For all $0\le x \le 1/2$ we have $(1-x)^{-A} \le (1+x)^{A} (1+x^2)^C$ for $C \ge \max\{-A,5A/3\}$. **Proof:** The inequality simplifies as $(1-x^2)^{A}(1+x^2)^C \ge 1$, or $f(t)=A\log(1-t)+C\log(1+t) \ge 0$ where $t \in I:=[0,1/4]$. Note $f(0)=0$ and that $f$ increases in $I$ if $C$ is large enough, namely $f'=-\frac{A}{1-t}+\frac{C}{1+t} = \frac{C-A-t(A+C)}{1-t^2} \ge 0$ for $t \le (C-A)/(C+A)$, which is $\ge 1/4$ for $C \ge \max\{-A,5A/3\}$.

By Claim 3 and the binomial series, $\sigma_p(f) = \sum_{n \ge 0}\binom{n+k-1}{n}^{\ell} p^{-n} \le \sum_{n \ge 0}\binom{n+k^{\ell}-1}{n} p^{-n} = \sum_{n \ge 0}\binom{n+k^{\ell}-1}{k^{\ell}-1} p^{-n} = (1-1/p)^{-k^{\ell}}$. This is small enough by Claim 4 with $x=1/p$, and we can take $c=\frac{5}{3}k^{\ell}$. $\blacksquare$

Regarding question 2: I do not know what the authors had in mind. I do suggest reading Selberg's elementary, and not sufficiently well known, for $\sum_{n \le x} \alpha(n) \sim Cx/\sqrt{\log x}$ where $\alpha$ is the indicator function of integers divisible only by primes of the form $4k+1$. This is essentially the claim you need. It is proved in 'Lectures on Sieves' in his collected papers, second volume, pages 183-185.

squarefreesums of two squares and then use the fact that, writing $n =m ^2 s$ with $s$ squarefree, $n$ is a sum of two squares if and only if $s$ is. $\endgroup$