Note that $Z \in (0, 5)$, so we have
$$ F_Z(Z \leq z) = \begin{cases} ? & z \leq 5 \\ 1 & z > 5 \end{cases} \\ $$
Now need to compute $F(Z \leq z | z \leq 5)$.
$$ F(Z \leq z | z \leq 5) = F(Z \leq z | Y > 5, z \leq 5)P(Y > 5, z \leq 5) + F(Z \leq z | Y < 5, z \leq 5)P(Y < 5, z \leq 5) \\ F(Z \leq z | Y > 5, z \leq 5) = F(X \leq z) = \frac{z}{5} \\ F(Z \leq z | Y < 5, z \leq 5) = 1 - P(X \geq z | Y < 5, z \leq 5)P(Y \geq z | Y < 5, z \leq 5) = 1 - (5 - z)^2/25 \\ F(Z \leq z | Z \leq 5) = z/10 + 1/2 - (5-z)^2/50 $$
So we have $$ F_Z(Z \leq z) = \begin{cases} z/10 + 1/2 - (5-z)^2/50 & z \leq 5 \\ 1 & z > 5 \end{cases} \\ \therefore f_Z(z) = \begin{cases} 1/10 + (5-z)/25 & 0 \leq z \leq 5 \\ 0 & \text{otherwise} \end{cases} $$
Is this the correct approach? Is there an easier approach to what I have taken here?
I suggest you a simple graphical approach:
Frist of all observe that
$$f_X(x)=\frac{1}{5}$$
$$f_Y(y)=\frac{1}{10}$$
in the following rectangle
Thus, by definition
$$\mathbb{P}[Z>z]=\mathbb{P}[X>z;Y>z]=\frac{(5-z)(10-z)}{50}$$
(area of the purple rectangle $\times f(x,y)$)
and thus
$$F_Z(z)=1-\frac{(5-z)(10-z)}{50}$$
derivating you get your density
$$f_Z(z)=\frac{15-2z}{50}\cdot\mathbb{1}_{(0;5)}(z)$$
... this result matches with yours but I think this procedure is very fast.