So cdf of $X$ is $F(X\leq x) = \frac{x}{\theta}$.
Let $X = \theta - Y$, subbing it into the cdf we now get: $$F(\theta - Y \leq x) = \frac{x}{\theta}$$ $$F(-Y \leq x - \theta) = \frac{x}{\theta}$$ $$F(Y\geq \theta - x) = \frac{x}{\theta}$$ $$F(Y\leq \theta - x) = 1 - \frac{x}{\theta}$$ $$=1 - F(X\leq x)$$
This what I came up with but no idea if this is right.
The CDF of $X$ for different cases is really
$$\begin{align*} \Pr(X \le x) = \begin{cases} 0 & x \le 0\\ \frac{x}{\theta}& 0 < x\le \theta\\ 1& \theta < x \end{cases} \end{align*}$$
Let $X=\theta - Y$, then the CDF of $Y$ is
$$\begin{align*} \Pr(Y \le y) &= \Pr(\theta - X \le y)\\ &= \Pr(X \ge \theta - y)\\ &= 1 - \Pr(X < \theta - y)\\ &= 1 -\begin{cases} 0 & \theta - y \le 0\\ \frac{\theta - y}{\theta}& 0 < \theta - y\le \theta\\ 1& \theta < \theta - y \end{cases}\\ &= \begin{cases} 0 & y < 0\\ \frac{y}{\theta}& 0 \le y < \theta\\ 1& \theta \le y \end{cases}\\ \end{align*}$$
Differentiate $\Pr(Y\le y)$ with respect to $y$ to obtain its PDF.