The sum of the floor function counts the integer points lying on or below a branch of the rectangular hyperbola $xy=n$ (for $x,y\gt 0$). Sketch this and add the square with corners $(0,0),(0,\sqrt n), (\sqrt n, 0), (\sqrt n, \sqrt n)$ (the last of these points lies on the hyperbola).
$k^2$ counts the points in the square, and the tails are symmetrical about $x=y$. The sum taken once counts the points in the tail, plus the points in the square (which are therefore counted twice when the sum is multiplied by $2$).
You should be able to turn these observations into a formal proof.
This is now the fourth time (at least) that a different form of question about this sum has been posed recently. The sum is equivalent to the sum of the divisor function $d(n)$, which counts the number of divisors of $n$.
Further note
If $(a,b)$ is below the hyperbola xy=n, so is $(b,a)$. Look at the sum on the right-hand side, taken once. It counts the number of points $(x,y)$ on the line $x=1$ with $0\lt xy \le n$ i.e. $\left\lfloor \frac n1 \right\rfloor$, the points with $x=2$ i.e. $\left\lfloor \frac n2 \right\rfloor$, up to the points with $x=k$ ie $\left\lfloor \frac nk \right\rfloor$.
Note that the points $(a,b)$ with $a,b \le k$ are all counted in this sum. No points with $a \gt k$ are counted.
The sum also counts the points in a similar way for $y=1, 2 \dots k$. This also has the points $(a,b)$ with $a,b\le k$ - we note there are $k^2$ of these, but has no points with $b\gt k$.
If we count the sum twice, we have counted all the points under the hyperbola, but there are $k^2$ points we have counted twice, so we subtract $k^2$ to avoid double-counting.
For other questions see this, and this.
Best Answer
Note that by the strict concavity of $\sqrt{x}$, Jensen's Inequality says $$ \hspace{-1cm}\sqrt{n}+\sqrt{n+1}+\sqrt{n+2}+\sqrt{n+3}+\sqrt{n+4}\lt5\sqrt{n+2}=\sqrt{25n+50}\tag{1} $$ More precisely, Taylor's Formula with remainder says $$ \begin{align} \sqrt{n+2+x} =\sqrt{n+2}+\frac{x}{2\sqrt{n+2}}-\int_0^x\frac{(x-t)\,\mathrm{d}t}{4\sqrt{n+2+t}^3}\tag{2} \end{align} $$ Summing $(2)$ for $x\in\{-2,-1,0,1,2\}$ yields $$ \hspace{-1cm}5\sqrt{n+2}-\frac5{4n^{3/2}}\lt\sqrt{n}+\sqrt{n+1}+\sqrt{n+2}+\sqrt{n+3}+\sqrt{n+4}\lt5\sqrt{n+2}\tag{3} $$ Since $$ \begin{align} \sqrt{25n+50}-\sqrt{25n+49} &=\frac1{\sqrt{25n+50}+\sqrt{25n+49}}\\ &\gt\frac1{2\sqrt{25n+50}}\\ &\gt\frac5{4n^{3/2}}\quad\text{for }n\ge14\tag{4} \end{align} $$ we get that for $n\ge14$, $$ \hspace{-5mm}\sqrt{25n+49}\lt\sqrt{n}+\sqrt{n+1}+\sqrt{n+2}+\sqrt{n+3}+\sqrt{n+4}\lt\sqrt{25n+50}\tag{5} $$ $(5)$ says that for $n\ge14$, $$ \left\lfloor\sqrt{n}+\sqrt{n+1}+\sqrt{n+2}+\sqrt{n+3}+\sqrt{n+4}\right\rfloor =\left\lfloor\sqrt{25n+49}\right\rfloor\tag{6} $$ It is simple to verify $(6)$ for $1\le n\lt14$ (it is false for $n=0$).