Approximation to $ \sqrt{2}$

Suppose $\frac{m}{n} $ is slightly bigger than $\sqrt{2}$, so that we can write $\frac{m}{n}= \sqrt{2}(1+\epsilon)$ where $\epsilon >0$ is small.

Then $$\frac{m+2n}{m+n} = \frac{ \frac{m}{n} +2}{\frac{m}{n} +1} = \frac{ \sqrt{2}(1+\epsilon)+2}{\sqrt{2}(1+\epsilon) + 1} = \sqrt{2} \left(1- \left(\frac{\sqrt{2}-1}{\sqrt{2}+1+\sqrt{2}\epsilon}\right)\epsilon \right)$$

Note that $\sqrt{2}+1+\sqrt{2}\epsilon> \sqrt{2}+1 $. Also, since $1<\sqrt{2}< \frac{3}{2}$, we have $\frac{\sqrt{2}-1}{\sqrt{2}+1} < \frac{1}{4}$ so ,$$\frac{\sqrt{2}-1}{\sqrt{2}+1+\sqrt{2}\epsilon}<\frac{1}{4}.$$

Thus, $\frac{m+2n}{m+n}$ is slightly smaller than $\sqrt{2}$ and it's difference from $\sqrt{2}$ is smaller in magnitude than the previous estimate, and decreases by at least a factor of $4$ with each iteration.

In a similar manner you can show the other case.

EDIT: I strengthened the estimates to address the rate of convergence issues Andres Caicedo brought up in a comment above.

HINT $\ $ Note that since $\rm\:\dfrac{m}n\:\dfrac{2\:n}m\:=\:2,\:$ one fraction is less than $\:\sqrt{2}\:$ and the other greater. Further their mediant $\rm\:\dfrac{m+2n}{n+m}\:$ is strictly between them, being the slope of the diagonal $\rm\:(n,m)+(m,2n)\:$ of the parallelogram formed by the vectors $\rm\:(n,m)\:$ and $\rm\:(m,2n)\:.\:$ To learn more search on the terms: mediant, Farey series and continued fraction.

We do the "opposite directions" and "better approximation" parts, including an estimate of how much better.

We are intended to assume that $m$ and $n$ are positive, and indeed that they are positive integers. In the argument below, we do not need $m$ and $n$ to be integers, but we do assume they are positive. Some assumption needs to be made, since $m=-1$, $n=1$ quickly leads to disaster!

Look at $$\frac{m+2n}{m+n}-\sqrt{2}.$$ This is equal to $$\frac{m+2n-m\sqrt{2}-n\sqrt{2}}{m+n},$$ which in turn is equal to $$-\frac{(\sqrt{2}-1)(m-n\sqrt{2})}{m+n}.$$ Divide top and bottom by $n$. We get that the above expression is equal to $$-\frac{(\sqrt{2}-1)(\frac{m}{n}-\sqrt{2})}{1+\frac{m}{n}}.$$ So we conclude that $$\frac{m+2n}{m+n}-\sqrt{2}=\left(-\frac{\sqrt{2}-1}{1+\frac{m}{n}}\right)\left(\frac{m}{n}-\sqrt{2}\right).$$

Note that the "multiplication factor" $-\frac{\sqrt{2}-1}{1+\frac{m}{n}}$ is negative. That means that if $\frac{m}{n}-\sqrt{2}$ is negative, then $\frac{m+2n}{m+n}-\sqrt{2}$ is positive, and if $\frac{m}{n}-\sqrt{2}$ is positive, then $\frac{m+2n}{m+n}-\sqrt{2}$ is negative. Thus the approximations alternate between too big and too small.

Note also that the multiplication factor $-\frac{\sqrt{2}-1}{1+\frac{m}{n}}$ has absolute value less than $\sqrt{2}-1$, which is less than $0.5$. So the absolute value of the error when we approximate $\sqrt{2}$ by $\frac{m+2n}{m+n}$ is less than half the absolute value of the error when we approximate $\sqrt{2}$ by $\frac{m}{n}$.

Note that we can make a better estimate of the rate of approach to $\sqrt{2}$, if we assume that we start with $m=n=1$. For then, forever, our approximation will be bigger than $1$, so the multiplication factor has absolute value $(\sqrt{2}-1)/(1+m/n)$, which is less than $(\sqrt{2}-1)/2$.