Using convolution:
$$ g_{X+Y}(t)=\int_\mathbb{R}g_X(t-x)g_Y(x)dx = \frac{1}{4}\int_{-1}^{1}\mathbb{1}_{[-1,1]}(t-x)dx = $$
Now, $t-x \in [-1, 1] \implies -x \in [-1-t,1-t] \implies x \in [t-1, t+1]$ and we may continue ($\vee$ stands for max, $\wedge$ for min):
$$ = \frac{1}{4} \int_{-1 \vee (t-1)}^{1 \wedge t+1}dx = \left\{ \begin{array}{ll} 0 & \textrm{$t \notin [-2,2]$}\\ \frac{t+2}{4} & \textrm{$t \in (-2, 0)$}\\ \frac{-t+2}{4} & \textrm{$t \in [0, 2)$} \end{array} \right. $$
Is that a correct answer? I had an example of X, Y being uniform, but on $[0,1]$ and extrapolated.
How can I calculate pdf of $X^2 + Y^2$?
I already know that $ F_{X^2}(t)=\mathbb{P}(X^2 \leq t) = \mathbb{P}(X \in [-\sqrt t, \sqrt t]) = \frac{\sqrt t - (- \sqrt t)}{1 - (-1)} = \sqrt t $ so $g_{X^2} = \frac{\partial F_{X^2}}{\partial t} = \frac {1}{2 \sqrt t} $
The joint density is given by $f(x,y)=\frac 1 4$ for $x,y \in [-1.1]$. So $P(X^{2}+Y^{2} \leq t)=\int_R f(x,y)dxdy$ where $R$ is the region defined by $x^{2}+y^{2} \leq t$. To evaluate this integral first integrate w.r.t $x$ from $-\sqrt {t-y^{2}}$ to $\sqrt {t-y^{2}}$ to and then integrate w.r.t. $x$ from $-\sqrt t$ to $\sqrt t$. You can use the substituion $y=\sin \theta$ to evaluate the integral.