Note that depending on how you orient your surface, then perhaps $\de\mb{S} \cdot \mb{k} = -\de{x}\de{y}$. I will think of the cone's normal as pointing 'outwards' and upwards, rather than 'inwards' and downwards. First, note that the situation looks like this:

The intersection of the plane and the cone can be verified to be given by the position vector $\mb{r} = [x, y, (1-x)/2]$ where $(x, y)$ satisfy the equation of an ellipse:
\[
\left(\frac{x + 1/3}{a}\right)^2 + \left(\frac{y}{b}\right)^2 = 1,
\]

where $a = 2/3$ and $b = 1/\sqrt{3}$. This curve is easily parameterized if we set $x = a\cos\theta$ and $y = b\sin\theta$. We wish to verify Stokes' Theorem:
\[
\int_{\partial S} \mb{F} \cdot \de{\mb{r}} = \iint_S \nabla \times \mb{F} \cdot \de{\mb{S}},
\]

where recall that $S$ needs to be a smooth, oriented surface, whose boundary is the curve $\partial S$, and is traversed in the positive sense. By 'positive', we mean that if we are walking along the boundary with the surface to our left, then the surfaces' normal is pointing (roughly) in the same reaction as our head. In the context of this problem, this means walking around the ellipse counter-clockwise.

We wish now to compute the line integral
\[
\int_{\partial S} \mb{F} \cdot \de{\mb{r}},
\]

where remember that for Stokes' Theorem, we need to be integrating along the contour in such a way so that the surface is to *left*. The position vector of the contour is then given by
\[
\gamma(\theta) = [x, y, z] \Bigr\rvert_{\partial S} = [a\cos\theta, b\sin\theta, (1-a\cos\theta)/2],
\]

and thus
\[
\gamma'(\theta) = [-a\sin\theta, b\cos\theta, \frac{a}{2}\sin\theta].
\]

To compute the line integral, we use
\begin{align*}
\int_{\partial S} \mb{F} \cdot \de{\mb{r}} &= \int_{2\pi}^0 \mb{F}(\gamma(\theta)) \cdot \gamma'(\theta) \, \de{\theta} \\
&= \int_{2\pi}^0 (ab\cos^2\theta) \, \de{\theta} \\
&= -ab\pi.
\end{align*}

Now we return to the surface integral. We parameterize the surface by
\[
\mb{r}(x, y) = [x, y, \sqrt{x^2 + y^2}],
\]

and compute the normal scaling vector,
\[
\left\lvert \pd{\mb{r}}{x} \times \pd{\mb{r}}{y}\right\rvert
= \left[ -\frac{x}{\sqrt{x^2 + y^2}}, -\frac{y}{\sqrt{x^2 + y^2}}, 1\right].
\]

However, note that this normal points inwards and downwards. Thus, we will reverse its direction in the computation of the surface integral, so that
\[
\de\mb{S} \cdot \mb{n} = -\de{x}\de{y}.
\]

We note also that
\[
\nabla \times \mb{F} = \mb{k},
\]

so the computation of the right hand side of Stokes' Theorem requires
\begin{align*}
\int_S (\nabla \times \mb{F}) \cdot \de{mb{S}} &= \int_S \mb{k} \cdot\de{\mb{S}} \\
&= \iint_{S_{xy}} (-1) \de{x}\de{y}.
\end{align*}

However, that last quantity is the surface area of an ellipse with semi-axes $a$ and $b$. Thus, by a formula (or a derivation),
\[
\int_S (\nabla \times \mb{F}) \cdot \de{mb{S}} = -\pi ab,
\]

verifying Stokes' Theorem.