I think I can show that
$$\sum_{1 \leq h,k \leq N} \frac{GCD(h,k)^2}{hk}$$
grows linearly. But I get the constant is
$$\sum_{ GCD(i,j)=1} \frac{1}{\max(i,j) i j}$$
This constant is incredibly close to $3$. (I am omitting the $1/12$, so my $3$ is your $0.25$.) My intuition is that they can't be equal, but they agree to a lot of digits, so I am not sure. See below for my computations.

UPDATE: This constant is $3$, due to an identity of Euler. See Marty's answer here. I'll leave the numeric work below for those might be curious how to approximate things like this.

Let's group the sum according to $GCD(h,k)$. So we have
$$\sum_d \sum_{\substack{1 \leq h,k \leq N \\ GCD(h,k)=d}} \frac{d^2}{hk} = \sum_d \sum_{\substack{1 \leq i,j \leq N/d \\ GCD(i,j)=1}} \frac{d^2}{d^2 ij} = \sum_d \sum_{\substack{1 \leq i,j \leq N/d \\ GCD(i,j)=1}} \frac{1}{ij}$$
where $h=di$ and $k=dj$. Grouping on $(i,j)$, we have
$$\sum_{\substack{1 \leq i,j \leq N \\ GCD(i,j)=1}} \frac{\lfloor N/\max(i,j) \rfloor}{ij} = N \sum_{\substack{1 \leq i,j \leq N \\ GCD(i,j)=1}} \frac{1}{\max(i,j) i j} + O \left( \sum_{\substack{1 \leq i,j \leq N \\ GCD(i,j)=1}} \frac{1}{i j} \right)$$
The error term is $O(\log N)^2$, so that's not the dominant term.

Once we check that the sum converges, this will show that your rate of growth is linear with that coefficient. We'll drop the $GCD(i,j)=1$ condition, since that just makes the sum smaller.
$$\sum_{i,j} \frac{1}{\max(i,j) i j} = \sum_{n} \frac{1}{n^2} \left( 2 + \frac{2}{2} + \frac{2}{3} + \cdots + \frac{2}{n-1} + \frac{1}{n} \right) = \sum_{n} n^{-2} O(\log n).$$
Here $n=\max(i,j)$. The final sum converges by the integral test, so the original one does as well.

Now, what is the value of this sum? Notice that, if $(i,j) = (g i', g j')$ with $GCD(i', j')=1$, then $\max(i,j)i j = g^3 \max(i',j') i' j'$. So, if we sum over all pairs, instead of just the relatively prime ones, then we multiply by a factor of $\sum g^{-3} = \zeta(3)$. So we want to compute
$$\sum_{1 \leq i,j} \frac{1}{\max(i,j) i j}$$
and, in particular, we want to know how it compares to $3 \zeta(3)$. As we showed above, we can simplify this sum to
$$\sum_{n} \frac{1}{n^2} \left( 2 + \frac{2}{2} + \frac{2}{3} + \cdots + \frac{2}{n-1} + \frac{1}{n} \right).$$

Now, is this actually the same as $3 \zeta(3)$? I had Mathematica compute the sum of the first $10,000$ terms, using $20$ digit precision for all intermediate computations. If Mathematica can be trusted, the result is $3.6040133$. Now, $2+2/2+2/3+\cdots +2/(n-1) + 1/n = 2 \log n + 2 \gamma + O(1/n)$. So I approximated the rest of the sum by the integral $\int_{10000}^{\infty} 2 (\log t + \gamma) dt/t^2$. (Here $\gamma$ is the Euler gamma constant.) According to Mathematica, this integral is $0.0021575$. The error in approximating a decreasing sum by an integral is bounded by the first term, which is $1.8 \times 10^{-7}$. The error in approximating the harmonic number by a $\log$ should be something like $\int_{10000}^{\infty} dt/t^3 = 5 \times 10^{-9}$; I don't have the energy to turn this into a rigorous bound. So the sum should be $3.6040133 + 0.0021575 \pm 2 \times 10^{-7} = 3.6061708 \pm 2 \times 10^{-7}$. (That error lines up with the last digit given.)

And what is $3 \zeta(3)$? I kid you not, it is $3.6061707$, right in range! So they might be equal, but, if so, I can't see why.

UPDATE: OK, I went back and improved my approximations in two ways: (1) I replaced the Harmonic number $H_n$ by $\log n + \gamma + (1/2) n^{-1} - (1/12) n^{-2} + (1/120) n^{-4}$ and (2) I approximated the sum of the terms past $10000$ by the first few terms of the Euler-Macluarin approximation, up to the $B_4$ term. The result, doing my internal computations with $20$ digits of accuracy: 3.6061707094787828562. The numerical value of $3 \zeta(3)$: Exactly the same!

Something is going on here. If you don't mind, I'll ask a separate question about what.