Even though i am not aware of a geometric interpretation of this formula, i think that the most explanatory answer to this is comparing it to the other Legendre's relation. Which states that for the set of Weierstrassian functions with periods $2\omega,2\omega'$ such that $Im(\omega'/\omega)>0$ and $\eta=\zeta(\omega), \eta'=\zeta(\omega')$, we have
$$\eta\omega'-\eta'\omega=\frac{\pi i}{2}.$$
This is in fact, equivalent to the elliptic integral identitiy. To prove this we integrate the quasi-elliptic function $\zeta$ around a period parallelogram. The total integral is sum of the residues inside the parallelogram, which is $1$ as there is only one pole. Hence
\begin{align}1 & =\frac{1}{2\pi i}\left( \int_{a}^{a+2\omega}+\int_{a+2\omega}^{a+2\omega+2\omega'}+\int_{a+2\omega+2\omega'}^{a+2\omega'}+\int_{a+2\omega'}^{a} \right)\zeta(u)du\\
&=\frac{1}{2\pi i}\int_{a}^{a+2\omega}\zeta(u)-\zeta(u+2\omega')du+\frac{1}{2\pi i}\int_{a}^{a+2\omega'}-\zeta(u)+\zeta(u+2\omega)du\\
&=\frac{2}{\pi i}\left( \eta\omega'-\eta'\omega \right)\end{align}
where $a$ is the base of the parallelogram and we are done. Now you might directly derive your form by the classical formulas
$$\omega=\frac{K}{\sqrt{e_1-e_3}},\qquad \omega'=\frac{iK'}{\sqrt{e_1-e_3}}$$
and
$$\eta=\sqrt{e_1-e_3}\left( E-\frac{e_1}{e_1-e_3}K \right),\qquad \eta'=-i\sqrt{e_1-e_3}\left( E'+\frac{e_3}{e_1-e_3}K' \right)$$
where $e_1=\wp(\omega), e_3=\wp(\omega')$.
This of course is not that sensible unless you know where the above formulas come from. But we can infact mimic the proof of other Legendre's relation. The Legendre identity may be stated as
$$J'K-JK'=\frac{\pi}{2}$$
with $J'=E',J=K-E$. This is very reminiscent of the above integral proof. We must look for an quasi-elliptic function with periods proportional to $K,K'$. The Jacobian $dn$ satisfies this with the periods $2K,4iK'$. So we might integrate $dn$ to obtain a function analogous to $\zeta$. This does not work however because $dn$ is an odd function and has simple poles. So consider $dn^2$, which has periods $2K,2iK'$. Let us denote
$$\zeta_{dn}(u)=\int_{0}^{u}dn^2udu.$$
Thus we expect there to be constants
$$\zeta_{dn}(u+2K)=\zeta_{dn}(u)+2\eta_{dn}, \qquad\zeta_{dn}(u+2iK')=\zeta_{dn}(u)+2\eta_{dn}'.$$
Evaluating $\eta_{dn},\eta_{dn}'$ and applying the procedure above, we obtain the Legendre's identity in terms of $J$. In fact, the function $\zeta_{dn}$ is Jacobi's form of the second elliptic integral $E(u)$ and it is no coincidence that it is also essentially equal to the so called Jacobian zeta function $Z$ given by
$$Z(u)=\int_{0}^{u}dn^2udu-\frac{E}{K}u=E(u)-\frac{E}{K}u=\zeta_{dn}(u)-\frac{E}{K}u.$$
Now let us elaborate these details. First of all, the actual definitions of $J$ and $J'$ in terms of elliptical integrals is given by
$$J=\int_{0}^{1}\frac{k^2t^2dt}{\sqrt{(1-t^2)(1-k^2t^2)}}, \qquad J'=\int_{1}^{1/k}\frac{k^2t^2dt}{\sqrt{(t^2-1)(1-k^2t^2)}}$$
following Weierstrass. Let us define a normalized zeta function
$$I(u)=\int_{0}^{u}k^2dn^2u du.$$
It is in fact true that
$$I(u+2K)=I(u)+2J, \qquad I(u+2iK')=I(u)+2iJ'$$
which implies Legendre's identity by the contour method. These relations can be derived simply by manipulation of integrals $I(K)=J$ and $I(iK')=iJ'$ since it is obvious that
$$I(u+2K)=I(u)+2I(K), \qquad I(u+2iK')=I(u)+2I(iK').$$
See below for example on how to obtain these identities.
There is also another direct method which highlights the relation and interplay between various Jacobian functions instead of using the powerful method of contour integration. By the standard techniques of integrating elliptic functions, it can be seen that there exists a constant $C$ such that
$$I(u)=Cu-\frac{\Theta'(u)}{\Theta(u)}$$
where $\Theta$ is one of the classic Jacobian theta functions
$$\Theta(u)=1-2q\cos\left( \frac{\pi u}{K} \right)+2q^4\cos\left( \frac{2\pi u}{K} \right)-2q^9\cos\left( \frac{3\pi u}{K} \right)+2q^{16}\cos\left( \frac{4\pi u}{K} \right)\cdots$$
with the nome $q=e^{-\pi K'/K}$. This is in fact the Jacobi's zeta identity $\Theta(u)=\Theta(0)\exp\left(\int_0^u Z(u)du\right)$ or $Z(u)=\Theta'(u)/\Theta(u)$. Putting $u=K$, we have $I(K)=CK$ since $\Theta'(K)=\Theta'(0)=0$. In the integral $I(K)$, let us put $t=sn (u)$ to obtain
$$I(K)=\int_{0}^{1}\frac{k^2t^2}{\sqrt{(1-t^2)(1-k^2t^2)}}dt=J$$
and thus $C=J/K$. In an analogous manner,
$$I(K+iK')=\frac{J}{K}(K+iK')-\frac{\Theta'(K+iK')}{\Theta(K+iK')}=\frac{J}{K}(K+iK')+\frac{\pi i}{2K}.$$
By the same subtitution and manipulation of elliptic integrals,
$$I(K+iK')=J+iJ'=\frac{J}{K}(K+iK')+\frac{\pi i}{2K}.$$
Which in turn implies $J'K-JK'=\pi/2$. Going back from the integral definition of $J,J'$, $J=K-E$ is obvious. The substitution $t\to\sqrt{\frac{1-k'^2t^2}{k^2}}$ shows that $J'=E'$.
For reference, you may see Weierstrass and Schwarz's Formeln und Lehrsätze zum Gebrauche der Elliptischen Funktionen. The Jacobian approach i have used here is based on the exposition of Hancock's Lectures on the Theory of Elliptic Functions, see article 249.
