asymptotic behavior of $\displaystyle{\sum_{r = 0}^k} { 2k - r \choose k} x^r $ for large $k$ Problem
This problem comes from the partition function $Z( n, r )$ in this question. I want to analyze the large $k$ behavior of
\begin{equation}
f(k)  = \sum_{r = 0}^k { 2k - r \choose k} x^r
\end{equation}
For example, 
\begin{equation}
\sum_{r =0}^k { 2k - r \choose k} 2^r = 4^k 
\end{equation}
A more specific question would be the limit as a function of $x$
\begin{equation}
\lim_{k\rightarrow \infty} \frac{\ln f(k)}{k}
\end{equation}
One can assume $x > 0$ or even $x > 1$. I'm interested at the point $x = 3$. 

A preliminary analysis shows that
\begin{equation}
\sum_{r=0}^{\infty} { 2k -r \choose k } x^r  = { 2k \choose k} {}_2F_1( 1, -2k;-k; x ) 
\end{equation}
then the problem reduces to the analysis of the Gauss hypergeometric function. 

Update 1: numerical data
I calculate the results for $x = 0.1$ to $10$ and $k$ up to $25$

$\ln f( x, k )$ vs $k$ is pretty much on the straight line even for $k$ of order $10$. 
Then I fit the curve (in range $k \in [15, 25]$) and compare the slope with the result from @ThomasAndrews

Based on this finite size numerical calculation, we can see that for $x> 4$, $2\ln x - \ln( x- 1)$ is a very accurate estimation. It is actually the exact answer( see the error estimation by @ThomasAndrews ). 
However, the error term can not be improved, because the data in range $[2,4]$ is slightly above $2\ln x - \ln( x- 1)$. The estimation doesn't work for $k < 2$. 

[2017-09-07]Update
Following @ThomasAndrews, we should analyze the scaling behavior of
\begin{equation}
f( k, x ) = x^k \sum_{r=0}^{k} { k + r \choose r } x^{-r}  \quad x \ge 2 
\end{equation}
I here present a slightly different break up. 
Using Pascal's rule for binomial coefficients
\begin{equation}
{ k + r \choose r} = { k - 1 \choose 0} + { k \choose 1 } + \cdots + { k+r - 1 \choose r } 
\end{equation}
we have
\begin{equation}
\begin{aligned}
f(k,x) = x^k\Bigg\{ &\left[ { k - 1 \choose 0} + { k \choose 1 } + \cdots + { 2k - 1 \choose k  } \right] \frac{1}{x^k} \\
 +&\left[ { k - 1 \choose 0} + { k \choose 1 } + \cdots + { 2k - 2 \choose k-1  } \right] \frac{1}{x^{k-1}}\\
 +\vdots &\\
+&\left[ { k - 1 \choose 0}\right] \frac{1}{x^0}  \Bigg\} \\
\end{aligned}
\end{equation}
Summing along the vertical lines to get a recursive relation 
\begin{equation}
\begin{aligned}
\frac{f(k,x)}{x^k}  &= \frac{x}{x - 1}\left[ (1 - x^{-k-1}){ k - 1 \choose 0} + x^{-1} ( 1 - x^{-k}){ k \choose 1 } + \cdots + x^{-k}( 1 -x^{-1} ){ 2k - 1 \choose k  } \right] \\
&=  \frac{x}{x - 1} \left[ \sum_{r =0 }^{k-1} { k - 1 + r \choose r } x^{-r}  + x^{-k}{2k - 1 \choose k }  - x^{-k-1} { 2k \choose k } \right]\\
&= \frac{x}{x-1} \frac{f( k-1, x ) }{x^{k-1}} + \frac{x - 2}{x-1} x^{-k}  { 2k - 1 \choose k } \quad k \ge 1\\
&= \left( \frac{x}{x - 1}\right)^k \left[ f(0,x) + \frac{x - 2}{x-1} \sum_{r=1}^{k} \left( \frac{x^2}{x - 1}\right)^{-r} { 2r - 1 \choose r } \right] 
\end{aligned}
\end{equation}
Therefore
\begin{equation}
f(k,x) = \left( \frac{x^2}{x - 1}\right)^k  + \frac{x - 2}{x-1} \left( \frac{x^2}{x - 1}\right)^k \sum_{r=1}^{k} \left( \frac{x^2}{x - 1}\right)^{-r} { 2r - 1 \choose r } 
\end{equation}
A consistency check is
\begin{equation}
f( k, 2 )  = 4^k \rightarrow \text{correct}
\end{equation}
For general $x > 1, x \ne 2$, we have $\frac{x^2}{x - 1} > 4$, which is in the convergence range of the series
\begin{equation}
\sum_{r=0}^{\infty} { 2r \choose r } x^r = \frac{1}{\sqrt{ 1- 4x} } \quad |x| < \frac{1}{4} 
\end{equation}
Consequently, we can use this generating function to approximate the second term of $f$ (the error is $o(1)$ because of the convergence of the series), 
\begin{equation}
\begin{aligned}
f( k ,x ) &=\left( \frac{x^2}{x - 1}\right)^k  + \frac{1}{2} \frac{x - 2}{x-1} \left( \frac{x^2}{x - 1}\right)^k \sum_{r=1}^{k} \left( \frac{x^2}{x - 1}\right)^{-r} { 2r \choose r } \\
&\sim  \left( \frac{x^2}{x - 1}\right)^k + \frac{1}{2} \frac{x - 2}{x-1} \left( \frac{x^2}{x - 1}\right)^k \left( \frac{x}{|x-2|}  - 1 \right) \\
&= \left( \frac{x^2}{x - 1}\right)^k + \frac{1}{x-1} \left( \frac{x^2}{x - 1}\right)^k  \quad x > 2 \\
&= \frac{1}{x} \left( \frac{x^2}{x - 1}\right)^{k+1}
\end{aligned}
\end{equation}
We conclude that
\begin{equation}
\label{eq:Z_N-asymp}
\lim_{k \rightarrow \infty} \frac{\ln f(k,x)}{k} = 2\ln x - \ln ( x+ 1 ) \quad x \ge 2
\end{equation}
I'm not sure why the fitting result is a bit off the theoretical line in $[2,4]$. 

Thanks for your attention. 

An upper bound, and a lower bound when $x>4$.
Assuming you mean $x^r$ rather than $x^{k}$, then you get:
$$\sum_{r=0}^{k}\binom{2k-r}{k}x^r = \sum_{s=0}^{k}\binom{k+s}{k}x^{k-s}=x^k\sum_{s=0}^{k}\binom{k+s}{k}x^{-s}$$
Now, $$\sum_{s=0}^{\infty}\binom{k+s}{k}y^s =\frac{1}{(1-y)^{k+1}}$$
for $|y|<1$.
So, when $x>1$, letting $y=\frac{1}{x}$, you get $f(k)\leq x^k\frac{1}{\left(1-\frac{1}{x}\right)^{k+1}}= \frac{x^{2k+1}}{(x-1)^{k+1}}$, or:
$\log f(k)\leq (2k+1)\log x - (k+1)\log(x-1)$ and thus:
$$\limsup_{k\to\infty}\frac{\log f(k)}{k}\leq 2\log(x)-\log(x-1)$$
The error function:
$$E_k=\frac{x^{2k+1}}{(x-1)^{k+1}}-f(k)=\sum_{s=1}^{\infty}\binom{2k+s}{k}x^{-s}$$
We can note that $\binom{2k+s}{k}<\binom{2(k+s)}{k+s}$ and get:
$$E_k<x^{k}\sum_{t=k+1}^{\infty} \binom{2t}{t}x^{-t}<x^{k}\left(\frac{1}{\sqrt{1-4/x}}-1\right)=\frac{4x^{k}}{\sqrt{x^2-4x}+x-4}=O(x^{k-1})$$ 
when $x>4$. 
I think that means, when $x>4$, that $\lim_{k\to\infty}\frac{\log f(k)}{k}=2\log(x)-\log(x-1)$.
