It is given that $\mathrm{f}(x) = x + \frac{m^2}{x-2}$, where $0 < m < 1$.

1. Identify the vertical and oblique asymptotes of the graph.

For $y = x + \frac{m^2}{x-2}$, the fraction is undefined when the denominator is zero.
As $x \to 2$, $y \to \pm\infty$, so the vertical asymptote is $x = 2$.
As $x \to \pm\infty$, $\frac{m^2}{x-2} \to 0$, so the curve approaches the line $y = x$.
The oblique asymptote is $y = x$.

2. Find the $x$-coordinates of the turning points by differentiating.

To find the turning points, set $\frac{\mathrm{d}y}{\mathrm{d}x} = 0$:
$\frac{\mathrm{d}y}{\mathrm{d}x} = 1 - \frac{m^2}{(x-2)^2}$
$1 - \frac{m^2}{(x-2)^2} = 0 \implies (x-2)^2 = m^2$
Since $m > 0$, taking the square root gives:
$x - 2 = \pm m \implies x = 2 - m \text{ or } x = 2 + m$

3. Calculate the corresponding $y$-coordinates of the turning points.

Substitute $x$ back into the original equation to find $y$:
When $x = 2 - m$:
$y = (2 - m) + \frac{m^2}{(2 - m) - 2} = 2 - m - m = 2 - 2m$
When $x = 2 + m$:
$y = (2 + m) + \frac{m^2}{(2 + m) - 2} = 2 + m + m = 2 + 2m$
The turning points are $(2-m, 2-2m)$ and $(2+m, 2+2m)$.

4. Deduce the sketch properties using the given condition $0 < m < 1$.

Since $0 < m < 1$, we have $2-m > 1 > 0$ and $2-2m > 0$.
This means the local maximum $(2-m, 2-2m)$ lies in the first quadrant, to the left of the asymptote $x = 2$.
The local minimum $(2+m, 2+2m)$ lies to the right of $x = 2$.
The $y$-intercept is at $y = -\frac{m^2}{2}$ (which is negative).
With these features, sketch the two branches bounded by $x=2$ and $y=x$.

1. Manipulate the given equation to relate it to the function $\mathrm{f}(x)$.

$x^2 - (2+k)x + (m^2 + 2k) = 0$
$x^2 - 2x + m^2 - kx + 2k = 0$
$x^2 - 2x + m^2 = k(x-2)$
Dividing by $(x-2)$ since $x=2$ is an asymptote and not a root:
$\frac{x^2 - 2x + m^2}{x-2} = k$
$x + \frac{m^2}{x-2} = k$
$\mathrm{f}(x) = k$
The roots of the equation correspond to the $x$-coordinates of the intersection points between the curve $y = \mathrm{f}(x)$ and the horizontal line $y = k$.

2. State the key features of the graph of $y = \mathrm{f}(x)$ derived from part (i).

From part (i), setting $\mathrm{f}'(x) = 1 - \frac{m^2}{(x-2)^2} = 0$ gives $x = 2-m$ and $x = 2+m$.
The local maximum point is $(2-m, 2-2m)$.
The local minimum point is $(2+m, 2+2m)$.
The $y$-intercept is found by setting $x=0$:
$\mathrm{f}(0) = 0 + \frac{m^2}{-2} = -\frac{m^2}{2}$

3. Determine the values of $k$ that give two positive roots on the left branch of the curve.

For the equation to have two distinct positive roots, the line $y=k$ must intersect the curve at two distinct points where $x>0$.
On the left branch ($x < 2$), the curve crosses the $y$-axis at $y = -\frac{m^2}{2}$, rises to the maximum at $y = 2-2m$, and then falls towards the vertical asymptote $x=2$.
To intersect this branch at two points with $x > 0$, the horizontal line $y = k$ must lie strictly above the $y$-intercept and below the maximum.
Thus, we require:
$-\frac{m^2}{2} < k < 2-2m$

4. Determine the values of $k$ that give two positive roots on the right branch of the curve and combine the results.

The right branch of the curve ($x > 2$) lies entirely in the region where $x$ is positive. It has a local minimum at $y = 2+2m$.
Any horizontal line $y = k$ strictly above this minimum will intersect the right branch at two distinct points, both of which will have $x > 2 > 0$.
Thus, we also have valid solutions when:
$k > 2+2m$
Combining these, the complete set of values is $-\frac{m^2}{2} < k < 2-2m \text{ or } k > 2+2m$.

1. Apply transformation $A$ to the original curve equation.

Transformation $A$ is a translation of $-2$ units in the $x$-direction, which means replacing $x$ with $x + 2$.
$y = \mathrm{f}(x+2) = (x+2) + \frac{m^2}{(x+2)-2} = x + 2 + \frac{m^2}{x}$

2. Apply transformations $B$ and $C$ to find the final equation.

Transformation $B$ is a stretch parallel to the $x$-axis with scale factor $\frac{1}{2}$, which requires replacing $x$ with $2x$.
$y = (2x) + 2 + \frac{m^2}{2x} = 2x + 2 + \frac{m^2}{2x}$

Transformation $C$ is a translation of $-2$ units in the $y$-direction, which means subtracting $2$ from the entire function.
$y = \left(2x + 2 + \frac{m^2}{2x}\right) - 2 = 2x + \frac{m^2}{2x}$

3. Compare the transformed equation with the given resulting equation to solve for $m$.

We are given that the resulting curve is $y = 2x + \frac{1}{8x}$. By comparing the terms with our deduced equation:
$\frac{m^2}{2x} = \frac{1}{8x}$
$m^2 = \frac{2}{8} = \frac{1}{4}$
Taking the square root:
$m = \pm \frac{1}{2}$
Since we are given $0 < m < 1$ in the question stem, we reject the negative root to obtain:
$m = \frac{1}{2}$