minimum travel from point to point with incremental steps
Interesting question. It is surprising that the answer depends only on $|x|+|y|$. For example, the same number of the steps is needed to reach $(1,1)$ or $(0,2)$.

The minimum step is the least non-negative integer $n$ such that $n(n+1)/2-(|x|+|y|)$ is even and nonnegative.
Here is an $O(1)$-time algorithm that returns the value described above, where least_n, i.e., $\left\lceil\frac{-1+\sqrt{8(|x|+|y|)+1}}2\right\rceil$, is the least nonnegative integer $n$ such that $n(n+1)/2-(|x|+|y|)\ge0$.
def minimum_steps(x,y):
    distance_to_origin := absolute_value(x) + absolute_value(y)
    least_n := ceiling((-1 + square_root(8 * distance_to_origin + 1)) / 2)
    gap := n * (n + 1) / 2 - distance_to_origin
    # 0 &lt;= gap &lt;= n - 1

    if gap is even:
        return least_n
    else if n is even:
        return least_n + 1
    else:
        return least_n + 2

Given the number of steps $s$, the distance travelled is $1+2+\cdots+s=s(s+1)/2$.

The correctness of the formula and algorithm above comes from the following characterization.
Proposition. The minimum count of steps from $(0,0)$ to $(x,y)$ is the least non-negative integer $n$ such that $n(n+1)/2-(|x|+|y|)$ is nonnegative and even.
Proof. If we can go to $(x,y)$ in $n$ steps, the sum of some numbers between 1 and $n$ or their negations must be $x$ and the sum of the remaining numbers or their negations must be $y$, i.e.,
$$\pm1\pm2\pm\cdots\pm n = |x| + |y|$$
for some choice of all $\pm$'s. That means,
$$1+2+\cdots+ n - (|x| + |y|)$$
is nonnegative and even.
Now it is enough to prove that $(x,y)$ is reachable in $n$ steps if $n(n+1)/2-(|x|+|y|)$ is nonnegative and even. Let us prove it by induction on $n$.

The base cases, $n=0$ or $1$ means $(x,y)=(0,0), (0,1), (1,0)$. These cases are immediate to verify.
Suppose, as induction hypothesis, it is correct for smaller $n$'s. Now consider the case of $n\ge2$ with $n(n+1)/2-(|x|+|y|)$ nonnegative and even. We can assume $x$ and $y$ are non-negative; otherwise, for example, if $x$ is negative, we can change $x$ to its absolute value and reverse the direction of all steps that are parallel to $X$-axis.
There are three cases.

$x \ge n$.  Let $x'= x-n$. Then $$(n-1)n/2-(|x'|+|y|)=n(n+1)/2-(|x|+|y|)$$ is nonnegative and even. By induction hypothesis, we can go to $(x',y)$ in $n-1$ steps. To reach $(x,y)$ in $n$ steps, we continue $n$ units in X-direction.
$y\ge n$. This is symmetric to the case above.
$0\le x\lt n$ and $0\le y\lt n$. There are two subcases.

$x\ge 2$ and $y\ge 2$. Let $x'=(n-1)-x$ and $y'=n-y$. Then $|x'|\le n-3$ and $|y'|\le n-2$. $$(n-2)(n-1)/2-(|x'|+|y'|)\ge(n-3)(n-4)/2\ge0.$$ The parity of $(n-1)(n-1)/2-(|x'|+|y'|)$ is the same as $n(n+1)/2-(|x|+|y|)$, i.e, even. By induction hypothesis, we can go to $(x',y')$ in $n-2$ steps . Reversing all the steps, we can go to $(-x', -y')$ in $n-2$ steps as well. To reach $(x,y)$ in $n$ steps, we can continue with two more steps, $n$ units in X-direction and $n+1$ units in Y-direction.
One of $x$ and $y$ is $0$ or $1$. Let $g(k)=k(k+1)/2-(|x|+|y|)$. The smallest nonnegative integer such that $g(k)\ge0$ is  $m=\left\lceil\frac{-1+\sqrt{8(|x|+|y|)+1}}2\right\rceil$. Since both $x$ and $y$ are $\lt n$ and one of them is $0$ or $1$, $$m\le \frac{1+\sqrt{8n+1}}2.$$
When $g(m)$ is even, $n=m$ by definition. When $g$ is odd, since either $g(m+1)=g(m)+(m+1)$ or $g(m+2)=g(m)+(m+1)+(m+2)$ must be even,  either $m+1$ or $m+2$ must be $n$. So, whether $g(m)$ is even or odd, we have $$n\le m+2.$$
Combing the two inequalities above, we have
$$n\le \frac{5+\sqrt{8n+1}}2,$$
which implies $n\le 6$.  Since X-direction and Y-direction are symmetric, let us assume $y=0,1$. Since $x\lt n$, $x\lt 6$. So it is enough to verify the cases where $(x,y)$ $\in $ $\{(0,0),(0,1),$ $(1,0),(1,1),$ $(2,0), (2,1),$ $(3,0), (3,1),$ $(4,0), (4,1),$ $(5,0), (5,1)\}.$ It is easy to verify each of them.



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