Why iterated limits are different from simultaneous limits?

For fixed $n$, we have $\lim_{m\rightarrow\infty}a_{m,n}=\lim_{m\rightarrow\infty}\dfrac{m}{m+n}=1$, so $\lim_{n\rightarrow\infty}\lim_{m\rightarrow\infty}a_{m,n}=\lim_{n\rightarrow\infty}1=1$.

The limit $\lim_{m,n\rightarrow\infty}a_{m,n}$ does not exist. If it were, then $\lim_{m,n\rightarrow\infty}a_{m,n}=\lim_{n\rightarrow\infty}\lim_{m\rightarrow\infty}a_{m,n}=\lim_{m\rightarrow\infty}\lim_{n\rightarrow\infty}a_{m,n}$, but in this case they are not equal.

$|a_{m,n}-L|<\epsilon$ for $m,n\geq N$ is the formalised meaning of $\lim_{m,n\rightarrow\infty}a_{m,n}=L$, in which case such $m,n$ vary freely from $N$, neither of which bounds the other, so this is in some sense that $m,n$ need no be comparable, they are independent, as @Rahul, @Arthur have noted.

Why do you assume the denominator will be about twice the numerator? That is like saying that $m\approx n$,which need not be true. Sure, you could calculate what the limit is when $m=n$, and it will be $\tfrac12$, as you can iterate limits, or you could think that $n$ grows twice at fast as $m$. As long as $m\to \infty$, then will $n\to \infty$ too, but for $m=100$ it will be $n=200$; if $m=500$ then $n=1000$; if $m=2\times 10^8$, then $n=4\times 10^8$... and so on.

Do the math for each pair of numbers and you'll see that they always yield $\tfrac13$ as result, so it seems fine to say that as $m,n\to\infty$ keeping the relation $n=2m$ the limit is $\tfrac13$, which you could have calculated checking that $$\lim\limits_{n=2m} \frac m {m+n}=\lim_m \frac m {m+2m}=\lim_m \frac m {3m}=\lim_m \frac 13=\frac13.$$

Even more: suppose that $n$ grows much faster than $m$, although both tend to $\infty$, say $n=m^2$ (so that when $m$ reaches $100$, $n$ will be $10 000$, for instance): what happens now? It would be reasonable to say that $n$ wins in the long run, so the denominator will be much bigger than the numerator and the limit will be zero: again.

Anyway, that's just an intuition, you can see that $$\lim\limits_{n=m^2} \frac m {m+n}=\lim_m \frac m {m+m^2}=\lim_m \frac 1 {1+m}=0.$$

The thing here is that by fixing a relation between $m$ and $n$ you are coming back to a limit with only one variable/index, and then there wouldn't be anything new. Although these limits with restrictions are useful many times, for doble sequences one wish to define a concept of limit that doesn't have to consider any particular path or relation. Well, in this case, that won't be possible, since changing the relation the limit also changes it's value.

But there are situations in which the double or simultaneous limit is well defined (and I haven't dealt with the definition of limit, just a few intuitive arguments). Consider $$\lim_{m,n} \frac {2m} {m^2+nm}=\lim_{m,n} \frac 2 {m+n}=\cdots$$ Since the numerator is constant, and the denominator will tend to infinity —no matter the 'relative velocities' of $m$ and $n$—, then the limit has to be zero (and it is what we would be able to prove using the definition of limit). You can even check that using the same relations as before, and iterating limits, all give also zero.

(IMPORTANT: But that last idea doesn't work backwards, not fully at least: you would never have enough examples of relations between $m$ and $n$ that make the limit go to zero, to be sure that the double limit is zero; there are infinite options and there could always be one you haven't tried and gives a different answer; and if that happened to be the case, what you can assert is that the double/simultaneous limit does not exist).

Simultaneous limit does not mean that $m$ and $n$ grow equally fast. It means that they grow independently. If, as they grow independently, there isn't one single limit that the expression tends to, then the simultaneous limit doesn't exist.