The topological Euler characteristic of a singular curve is not always even. In fact, if $\tilde{C}$ is the normalization of $C$, then $\chi(C)=\chi(\tilde C)-n$ where $n$ is the number of nodal points. Indeed, let $x$ be a nodal point of $C$ and $C_x$ be the blow-up of $C$ at $x$. Then, the cartesian and cocartesian square
$$\require{AMScd}
\begin{CD}
\{x_1,x_2\}@>>> C_x\\
@VVV@VVV\\
x@>>>C
\end{CD}
$$
gives a long exact sequence in cohomology :
$$ \dots\rightarrow H^n(C)\rightarrow H^n(C_x)\oplus H^n(x)\rightarrow H^n(\{x_1,x_2\})\rightarrow H^{n+1}(C)\rightarrow\dots $$
In particular $\chi(C)+\chi(\{x_1,x_2\})=\chi(\{x\})+\chi(C_x)$. And because the characteristic of a point is $1$, $\chi(C)=\chi(C_x)-1$. By induction on the number of nodes, you get $\chi(C)=\chi(\tilde{C})-n$.
Here is the statement of the Riemann-Hurwitz Theorem I alluded to in the comments. (Rosen, Number Theory in Function Fields, Theorem 7.16, p. 90).
Theorem. Let $L/K$ be a finite, separable, geometric extension of function fields. Then
$$
\DeclareMathOperator{\D}{\mathfrak{D}}
\DeclareMathOperator{\P}{\mathfrak{P}}
2g_L - 2 = [L:K] (2g_K - 2) + \deg_L(\D_{L/K})
$$
where $\mathfrak{D}_{L/K}$ is the different ideal.
If all the ramified primes of $L$ are tamely ramified (which is the case here since the ground field has characteristic $0$), then $\D_{L/K} = \sum_{\P} (e(\P/P) - 1) \P$, so the formula becomes
$$
2g_L - 2 = [L:K] (2g_K - 2) + \sum_{\P} (e(\P/P) - 1) \deg_L(\P) \, .
$$
Turning to your example, your mistake is that $F$ is not ramified above $\infty$. A geometric way to see this is the following. Homogenizing the curve defining $F$, we obtain the curve $C: X^2 - (Y^2 - 10YZ - 5Z^2) = 0$, where $x = X/Z$ and $t = Y/Z$, and we are considering the map $\pi: C \to \mathbb{P}^1$, $[X:Y:Z] \mapsto [Y:Z]$. To compute $\pi^{-1}([1:0])$, we plug $Z = 0$ into the equation for $C$, obtaining
$$
0 = X^2 - Y^2 = (X-Y)(X+Y)
$$
so $\pi^{-1}([1:0]) = \{[1:1:0], [1:-1:0]\}$. Since $\sum_i e_i f_i = 2$ by the fundamental identity, then $f_i = e_i = 1$, so $\pi$ is unramified above $\infty$.
For a more function field theoretic approach, let $s = 1/t$ and $r = x/t = xs$. Then maximal order of $F$ at infinity is $R := \frac{\mathbb{Q}[r,s]}{(r^2 - (1 - 10s - 5s^2))}$. To determine the splitting above $\infty$, we examine how $sR$ factors. Using the equation defining $R$, we find $sR = (r-1,s)(r+1,s)$, and these primes are distinct, so $F$ is unramified above $(s)$.
Let $\P = (x)$ and $P = (t^2 - 10t - 5)$. The residue field of $\P = (x)$ is
\begin{align*}
\frac{\mathbb{Q}[t,x]/(x^2 - (t^2 - 10t - 5))}{(x)} \cong \frac{\mathbb{Q}[t]}{(t^2 - 10t - 5)}
\end{align*}
which has dimension $2$ as a $\mathbb{Q}$-vector space, so $\deg_L(\P) = 2$.
Applying Riemann-Hurwitz, we have
\begin{align*}
2g_L - 2 = 2 (2 \cdot 0 - 2) + (e(\P/P) - 1) \deg_L(\P) = -4 + (2-1) \cdot 2 = -2
\end{align*}
so $g_L = 0$, as we had hoped.
Best Answer
A normalization induces a bijection of irreducible components, which means your conclusion that "$\widetilde{C}$ has three components" is incorrect. $\widetilde{C}$ has two components, in fact, and it's just the disjoint union of the cubic and the line. The relevant marked points are just the preimages of the nodes, so we have a disjoint union of a curve of genus one and a curve of genus zero, both with three marked points, and thus our curve is stable.