Functions that gradually slow down as X gets bigger

If $x$ is the horizontal coordinate in your first graphic, the corresponding value is simply $f(x)=-\frac 12+\frac 12 \sqrt{1+8x}$.

For the second graphic you can use $g(x) = -\frac 32 +\frac 12 \sqrt{9+16x}$.

You can obtain these functions just by fitting the values with a second degree polynomial.

I'll elaborate on @PierreCarre's answer. What we need is a function $f$ that takes a number of steps $s$ and tells us what milestone we have reached ($f(s)$). I will first derive its inverse, because that's where my natural exploration brought me.

Going from the first milestone to the second requires one step. More generally, starting from the $n$th milestone, one would need $n$ steps to reach the next milestone (the $(n+1)$th one). Hence, to reach the $(n+1)$th milestone from the first, one would need to sum all these steps: $$\sum_{i=1}^ni = 1 + 2 + \dots + n.$$ There exists a closed formula for this sum (famously derived by Gauss): $$\sum_{i=1}^ni = \frac{n(n+1)}2.$$ This formula asks us what milestone, and tells us the number of steps required. You can check that this is the inverse of $f$ by comparing the formulations. Now remains to find $f$. Let $$g(n) = \frac{n(n+1)}2.$$ Now let's rewrite: $$\iff 2g(n) = n^2 + n$$ $$\iff n^2 + n -2g(n) = 0.$$ We can use the quadratic formula, using $n$ as the unknown variable, to find that $$n = \frac{\sqrt{8g(n) + 1} - 1}2 \text{ or } n = \frac{-\sqrt{8g(n) + 1} - 1}2.$$ Now notice that $g(n)$ and $s$ both denote an amount of steps needed to reach a certain milestone. Similarly, notice that $f(s)$ and $n$ both denote a milestone. We can hence rewrite the above formulas as follows: $$f(s) = \frac{\sqrt{8s + 1} - 1}2 \text{ or } f(s) = \frac{-\sqrt{8s + 1} - 1}2.$$ The latter is a decreasing function and hence not what we want. The former corresponds to the formula in the other answer, and it seems to pass the sanity check of inserting the first few numbers.

As for the second formula, you are summing $$1 + 1.5 + 2 + 2.5 + ... + x = \frac12(2+3+4+...+n),$$ where $x$ is an integer $n$ divided by two. We recognise the sum of the first $n$ integers between the parentheses, except the first term ($1$) is omitted. We can again use Gauss' formula to rewrite this, keeping the subtracted $1$ in mind. $$\dots = \frac12\left((1+2+3+4+...+n) - 1\right) = \frac12\left(\frac{n(n+1)}2 - 1\right).$$ It remains to find the inverse of this formula.