Solving $$f(x, k) = f(x, k-1) + f(x-1, k-1) + \dots + f(1, k-1)$$ in terms of $x$
The answers to all three questions become easier once we proved the following formula for all positive integer $k$ and $x$.
$$ f(x,k) = \frac{x(x+1)(x+2)\cdots(x+k-1)}{k!}$$
Proof: The formula is easily seen to be true for $k=1$ and all positive integer $x$.

Assume the formula is true for some positive integer $k$ and all positive integer $x$.
Then, 
$$\begin{align}
 &amp;f(x, k+1)\\
&amp;=f(x, k) + f(x-1, k) + f(x-2, k) + \dots + f(1, k)\\
&amp;=\frac1{k!}(x(x+1)(x+2)\cdots(x+k-1) + (x-1)x(x+1)(x+2)\cdots(x+k-2)\\ 
&amp;\quad\quad+\cdots+1\cdot2\cdots k\\
&amp;=\frac1{k!}\left(\frac{x(x+1)(x+2)\cdots(x+k)-(x-1)x(x+1)\cdots(x+k-1)}{k+1}\right.\\
&amp;\quad\quad+\frac{(x-1)x(x+1)\cdots(x+k-1)-(x-2)(x-1)x\cdots(x+k-2)}{k+1}\\
&amp;\quad\quad+\cdots\\
&amp;\quad\quad+\frac{2\cdot3\cdots(k+1)-1\cdot2\cdots k}{k+1}\\
&amp;\quad\quad+\left.\frac{1\cdot2\cdots k}{k+1}\right)\\
&amp;=\frac1{k!}\frac{x(x+1)(x+2)\cdots(x+k)}{k+1}\\
&amp;=\frac{x(x+1)(x+2)\cdots(x+(k+1)-1)}{(k+1)!}\end{align}$$

By mathematical induction on $k$, the formula is true for all positive integer $k$ and $x$.



Here are the answers to your three cases.

1) For fixed $k$ and $x\to\infty$, 
$$f(x,k)=\frac{x^k}{k!}\frac xx\frac{x+1}x\cdots\frac{x+k-1}{x}\sim\frac{x^k}{k!}$$
That is, $f(x,k)=\Omega(x^k)$.

2) For fixed $x$ and $k\to\infty$, by Stirling's approximation,
$$\begin{align}f(x,k)&amp;=\frac{(x-1+k)!}{(x-1)!k!}\\
&amp;\sim\frac1{(x-1)!}\frac{\sqrt{2\pi(x-1+k)}\left(\frac{x-1+k}{e}\right)^{x-1+k}}{\sqrt{2\pi k}\left(\frac{k}{e}\right)^{k}}\\
&amp;=\frac1{(x-1)!}\sqrt\frac{x-1+k}k
(x-1+k)^{x-1}\left(\frac{x-1+k}{k}\right)^k\frac{1}{e^{x-1}}\\
&amp;=\frac1{(x-1)!}\sqrt\frac{x-1+k}k\left(\frac{x-1+k}{k}\right)^{x-1}k^{x-1}\left(1+\frac{1}{\frac k{x-1}}\right)^{\frac k{x-1}(x-1)}\frac{1}{e^{x-1}}\\
&amp;\sim\frac1{(x-1)!}k^{x-1}e^{x-1}\frac{1}{e^{x-1}}\\
&amp;=\frac{k^{x-1}}{(x-1)!}\end{align}$$
That is, $f(x,k)=\Omega(k^{x-1})$

3) Let $k=x-1$ go to infinity, again by Stirling's approximation,
$$f(x,k) = \frac{(2k)!}{k!k!}\sim \frac{\sqrt{2\pi2k}\left(\frac{2k}{e}\right)^{2k}}{\sqrt{2\pi k}\left(\frac{k}{e}\right)^{k}\sqrt{2\pi k}\left(\frac{k}{e}\right)^{k}}=\frac{2^{2k}}{\sqrt{\pi k}}$$
That is, $f(k+1,k)=\Omega(k^{-1/2}2^{2k})$.