Quoting from the posted information for finding $P(W \le w):$
When $0 < w \le 1,$ the region $R_w$ is a triangle with vertices
$(0,0), (0, w),$ and $(w, 0).$ Then this triangle has area $w^2/2.$
Because the distribution of $X$ and $Y$ is uniform on the square
with vertices $(0,1)$ and $(1,1),$ probabilities are proportional to areas.
Thus $P(W \le w) = w^2/2,$ for $0 < w \le 1.$
If you want the density function, then differentiate the CDF, first
for $0 < w < 1$ and then for $1 <w \le 2.$
A simulation of 50,000 values of $W$ makes it easy to show some areas, the
PDF and the CDF. In the figure below, the first plot shows (red) the area corresponding to $P(X < 0.5);$ the second shows (green) the area corresponding
to $P(W < 1.6);$ the third shows the empirical CDF, which suggests the
shape of the CDF of $W$ you are asked to compute; and the fourth shows a histogram
of simulated values of $W.$ which suggests the triangular shape of the
density function of $W.$
The empirical CDF (ECDF) jumps up by $1/50,000$ at each observed value of $W.$
Usually, an ECDF gives a more precise indication of the CDF than a histogram gives of the PDF (because information is lost when data are put into histogram bins).
Addendum: R code for figure per Comments.
par(mfrow=c(2,2)) # enables 4 panels per plot
m = 50000; x = runif(m); y = runif(m); w = x+y
plot(x, y, pch=".", main="Joint Dist'n of X and Y with P(W < .5)")
points(x[w < .5], y[w < .5], pch=".", col="red") # read  as 'such that'
plot(x, y, pch=".", main="Joint Dist'n of X and Y with P(W < 1.6)")
points(x[w < 1.6], y[w < 1.6], pch=".", col="green2")
plot(ecdf(w), main="Emplrical CDF of Observed Values of W") # 'ecdf' procedure
hist(w, prob=T, col="skyblue2", main="Histogram of Observed Values of W")
points puts points onto an existing plot.
ecdf includes data sort and and cumulative tally, then returns x and y values ready for plotting (actually a stair-step plot but seems a curve with 50,000 points at this resolution); argument
hist function invokes vertical density scale.