The short answer: The "curved line" is not a straight line.
The longer answer: It may be useful to have an explicit situation, where we add reduced divisors. Let us consider the Example from the linked wiki page.
$$
(C)\ :\qquad
y^2 =x(x+1)(x+3)(x-3)(x-5)\ .
$$
Consider the points $P,Q,R,S$, and the reduced divisors $D_1,D_2$:
$$
\begin{aligned}
P &= (1, 8)\ , \\
Q &= (3, 0)\ ,\\
R &= (5, 0)\ ,\\
S &= (0, 0)\ ,\\[3mm]
D_1 &= P+Q-2O\ ,&&\text{Mumford representation:} & ((x-1)(x-3),&\ -4(x-3)\ ,\\
D_2 &= R+S-2O\ ,&&\text{Mumford representation:} & (x(x-5),&\ 0)\ .
\end{aligned}
$$
(There is no $[P]$ notation to pass from geometry, a point $P$, to a linearized object, $[P]$, a divisor. It should be clear from the context.)
It turns out that the sum $D=D_1+D_2$ is the divisor with the Mumford representation
$$
(x^2 - 8x+3, \ -12x)\ .
$$
So if we want to represent it as $T+U-2O$, we have to take the roots of $x^2-8x+3$, which are $4\pm s$ with $s=\sqrt {13}$ for an easy typing, and so we get the points $T,U$ equal to $(4\pm s,\ 12(4\pm s))$.
Details follow.
The cubic that passes through $P, Q, R, S$, is
$$
a(x) := x(x-3)(x-5)\ .
$$
Consider now the cubic $(C')$ with equation:
$$
(C')\ :\qquad y = a(x) = x(x-3)(x-5)\ .
$$
We intersect now the cubic $(C')$, $y=a(x)$, with the given hyperelliptic curve of genus two, $y^2 =f(x)$. Intersecting means solving the system with two equations, those for $(C')$ and $(C)$, with two unknowns ($x,y$, the ones that parametrize our affine piece of geometry):
$$
\left\{
\begin{aligned}
y &= a(x)\ ,\\
y^2 &= f(x)\ .
\end{aligned}
\right.
$$
How many solutions do we have?
This is more or less your question. (I found it easier to understand the situation in a sample case like the chosen one.)
Well, we take $y=a(x)$ from the one equation, substitute it in the second one, and obtain one equation in $x$, which is $a(x)^2 =f(x)$, in our case:
$$
x^2(x-3)^2(x-5)^2 = x(x+1)(x+3)(x-3)(x-5)\ .
$$
We move everything on the L.H.S. - factors are transparent, and we know that
$P,Q,R,S$ (i.e. their $x$-component) should be among the roots:
$$
\begin{aligned}
a(x)^2 - f(x)
&= x^2(x-3)^2(x-5)^2 - x(x+1)(x+3)(x-3)(x-5)
\\
&= x(x-3)(x-5)\Big( (x^3 - 8x^2 + 15x)- (x^2 + 4x+3)\Big)
\\
&=x(x-3)(x-5)\cdot(x^3 -9x^2 +11x-3)\qquad\text{(and $(x-1)$ from $P$ is a factor)}
\\
&=x(x-3)(x-5)\cdot(x-1)(x^2-8x+3)\ .
\end{aligned}
$$
A plot of $(C)$ and $(C')$, with expected shapes, and their intersection points follows.
(The points $P,Q,R,S$ we start with are green, the "new" points $T,U$ are painted magenta.)
Sage code producing the figure:
K.<s> = QuadraticField(13) # makes s represent sqrt(13)
r.<x> = K[]
f = x(x + 1)(x + 3)(x - 3)(x - 5)
C = HyperellipticCurve(f)
P = C(1, 8)
Q = C(3, 0)
R = C(5, 0)
S = C(0, 0)
T = C((4 + s, 48 + 12s))
U = C((4 - s, 48 - 12s))
var('y');
g = implicit_plot(y^2 - f, (x,-10, 20), (y, -150, 150)
, aspect_ratio=1/10
, figsize=(10, 10))
g += implicit_plot(y - x(x - 3)(x - 5), (x,-10, 20), (y, -150, 150)
, aspect_ratio=1/10
, figsize=(10, 10)
, color='brown')
g += point([(1, 8), (3, 0), (5, 0), (0, 0)]
, size=30, color='green')
g += point([(4 + s, 48 + 12s), (4 - s, 48 - 12s)]
, size=30, color='magenta')
And let now show $g$ with the default viewer... (Type g for instance in the interpreter.)
