Take a look at the junction of the two springs. If you displace the block and hold it there, the junction will also be at rest.
At the junction, two forces are pulling it, $k_1 x_1$ to the left and $k_2 x_2$ to the right.
Since the junction is at rest, the two forces must balance. Therefore, $$k_1 x_1 = k_2 x_2$$
EDIT 1:
Another calculus-based derivation which I feel is more fundamental:
For a given displacement of the block, the springs will distribute the total elongation between themselves in such a way that the total potential energy of the system is minimum. $$P.E=\frac 12 k_1 {x_1}^2+\frac 12 k_2 {x_2}^2$$
Since the potential energy is to be minimum for a given displacement of the block, we take its derivative with respect to any one of the elongations (say $x_1$) and equate it to zero. (You can always check whether it is actually a minima by doing the second derivative test).
$$\frac {d(P.E)}{d{x_1}}=\frac 12 k_1 (2x_1)\frac {dx_1}{dx_1}+\frac 12 k_2 (2x_2)\frac {dx_2}{dx_1}=0$$ $$\implies k_1 x_1 + k_2 x_2 \frac {dx_2}{dx_1} = 0$$
As $x_1 + x_2=x \implies x_2=x-x_1 $ (where $x$ is the given displacement of the block),$$k_1 x_1 + k_2 x_2 \frac {d(x-x_1)}{dx_1} = 0$$ $$\quad \implies k_1 x_1 + k_2 x_2 (0-\frac {dx_1}{dx_1}) = 0\quad $$ $$\quad \implies k_1 x_1 + k_2 x_2 (-1) = 0\quad $$ $$ \implies \boxed {k_1 x_1 = k_2 x_2} $$