The purpose of this answer is to show how a direct application of convolution may lead to the desired result. I take the following results from Cohn, Measure Theory.

**Definition of convolution** Let $\nu_1$ and $\nu_2$ be finite measures on $(\mathbb{R}^d,\mathscr{B}(\mathbb{R}^d))$, then their *convolution* $\nu_1\ast\nu_2$ is defined by:

$$ \nu_1 \ast\nu_2(A) = \nu_1 \times\nu_2(\{(x_1,x_2) : x_1+x_2 \in A\})$$

**Proposition 10.1.12** Let $\nu_1$ and $\nu_2$ be probability measures on $(\mathbb{R}^d,\mathscr{B}(\mathbb{R}^d))$.

$\vdots$

(c) If $\nu_1$ and $\nu_2$ are absolutely continuous, with densities $f$ and $g$, then $\nu_1\ast\nu_2$ is absolutely continuous with density:
$$x \mapsto \int f(x-y)g(y)\lambda(dy)$$

Let $I$ denote the unit interval $[0,1]$, and $U(I)$ the uniform distrbution on $I$. Then the density function corresponding to $U(I)$ is $\chi_I$, the indicator function for $I$. If $X$ and $Y$ are independent random variables whose distributions are given by $U(I)$, then the density of their sum is given by the convolution of their distributions. I.e., if $f_X$ denotes the density for random variable $X$, then

$$ f_{X+Y}(x) = \int f_X(x-y)f_Y(y)\lambda(dy) = \int \chi_I(x-y)\chi_I(y) dy$$

The indicator function of $y$ alone restricts the integration range, so that

$$ \int \chi_I(x-y)\chi_I(y)dy = \int_0^1 \chi_I(x-y) dy$$

The expression $\chi_I(x-y)$ is $0$ if $x-y < 0$ or $x-y > 1$:

$$\chi_I(x-y) = \cases{1 & $x-1 \leq y \leq x$ \\ 0 & otw} $$

This further restricts the range of the integral, which can be rewritten:

$$\int_{max(0,x-1)}^{min(1,x)} 1 dy = min(1,x) - max(0,x-1)$$

The density is $0$ if $x < 0$ or $x > 2$. This fact is hidden in our final expression because we've expressed our indicator functions through the bounds of the integral, but can be recovered by including another indicator function. The PDF as described in the original question follows by considering the relevant cases.