Taken care of it myself:
Since the proposed solution is a function of $Y$ we don't need to check measurability constraints for whether the solution is a version of conditional. For the integration part, It would be enough to check the integration inequalities for $C=Y^{-1}([0,c])$ for any $c\in[0,\theta]$. Moreover it is only necessary to check the integration for $B$'s in Borel $\sigma$-field of $[0,\theta]$ such that $B=[0,b]$. The discussed integration looks as follows:
$$ \int_C \frac{1}{2} I_B(Y)dP + \int_C\frac{1}{2}\frac{\lambda(B\cap [0, Y])}{Y}dP \overset{?}{=}\int_C I_B(X_1)dP.$$
We consider two cases $b\leq c$ and $c< b$. In the first case we have:
$$
\int_{Y^{-1}([0,c])}\frac{1}{2}I_B(Y)dP = \int_{Y^{-1}([0,b])}\frac{1}{2}I_B(Y)dP=\int_{0}^{b}\int_{0}^b \frac{1}{2\theta^2} dx_1dx_2=\frac{b^2}{2\theta}.
(*)
$$
Notice that if $c<b$ we get:
$$
\int_{Y^{-1}([0,c])}\frac{1}{2}I_B(Y)dP = \frac{c^2}{2\theta^2}.
(**)
$$
Now for the right term of the equation we have:
$$
\int_{Y^{-1}([0,c])} I_B(X_1)dP = \int_{0}^{c}\int_{0}^{b}\frac{1}{\theta^2}dx_1dx_2=\frac{bc}{\theta^2}.
(\*\*\*)
$$
Once again if $c<b$ we get:
$$
\int_{Y^{-1}([0,c])} I_B(X_1)dP = \frac{c^2}{\theta^2}
(\*\*\*\*)
$$
Finally
$$
\int_{Y^{-1}([0,c])}\frac{1}{2}\frac{\lambda(B\cap[0,Y])}{Y}dP=\int_{X_1\leq X_2 \leq c}\frac{1}{2}\frac{\lambda(B\cap[0,X_2])}{X_2}dP +\\
\int_{X_2\leq X_1 \leq c}\frac{1}{2}\frac{\lambda(B\cap[0,X_1])}{X_1}dP=\int_{X_1\leq X_2 \leq c}\frac{\lambda(B\cap[0,X_2])}{X_2}dP
$$
Notice that the final equation follows from the symmetry. Then we have:
$$
\int_{X_1\leq X_2 \leq c}\frac{\lambda(B\cap[0,X_2])}{X_2}dP=\int_{X_1\leq X_2\leq b\leq c} dx_1dx_2+\\
\int_{X_1\leq b\leq X_2\leq c} \frac{b}{x_2}\frac{1}{\theta^2}dx_1dx_2+\int_{b\leq x_1\leq x_2\leq c}\frac{1}{\theta^2}\frac{b}{x_2}dx_1 dx_2=\\
\frac{1}{\theta^2}\left[\int_{0}^{b}(b-x_1)dx_1 + \int_{0}^{b}b\ln(x_2)\Big|_b^c dx_1+ \int_{b}^{c} b\ln(x_2)\Big|_{x_1}^{c}dx_1\right]=\\
\frac{1}{\theta^2}\left[\frac{b^2}{2}+b^2\ln(c) - b^2\ln(b)+(c-b)b\ln(c)-bc\ln(c)+bc+b^2\ln(b)-b^2\right]=\frac{1}{\theta^2}\left(bc-\frac{b^2}{2}\right)
$$
which completes the proof for the case $b\leq c$ considering (*) and (***). For the other case, i.e. $c<b$ we have:
$$
\int_{Y^{-1}([0,c])}\frac{1}{2}\frac{\lambda(B\cap[0,Y])}{Y}dP=\int_{X_1\leq X_2\leq c\leq b} \frac{1}{\theta^2}dx_1 dx_2=\int_{0}^{c}(c-x_1)dx_1=\frac{c^2}{2\theta^2}
$$
which completes the proof for the $c<b$ case too, taking into account (**) and (****).