Due to symmetry in x,y we can do 45 deg rotation to bring "ellipse" axes along $ x$ and $y$. Let use rotationally transform $ x_1 = x-y, y_1 = x+y $ and ignore scaling of axes.
The contour is of a topography of hills and valleys.
Near to "Col" points ( flatter place to rest during mountaineering) between "ellipse" centers level curves more hyperbolic with saddle points.
At higher altitudes they are more elliptic and intersection ovals appear to be ellipses but indeed they are not so.
Ellipses are not periodic, meaning they have a one time appearance in the entire $x,y$ interval $ -\infty< x < \infty, -\infty< y < \infty $ of dimension 2.
But the given curves are an infinite degree polynomial trigonometric array whose dimension cannot be two as it is for a conic section.
This fact alone justifies to its instant recognition to being other than a conic section.
From the exact equation you have given a second degree approximation around its center can lead you to an ellipse, as per the following second degree approximation.
$ \cos x + \cos y = \pm \cos (x+y) $ , including negative sign if you shift
attention to the hyperboloid side of contour as well.
$ \cos x \approx. 1 - x^2/2 ,\cos y \approx 1 - y^2/2 ,\cos (x+y) \approx 1 - (x+y)^2/2 $ respectively simplify to
$ x^2 + y^2 + x\, y =1 $ and $ x\, y + 1 =0 $ . By taking 3,4,5 number of terms the 'ellipse' or 'hyperbola' can be approximated. So what you now see are a higher order ellipses/hyperbolas.
With respect to the oval center it looks like an ellipse. With respect to a center point of ovals as center you notice that..
the hyperbola look-like profile is ignored as, perhaps people look at the more round profile only. Different viewpoint centers are attached in the last illustration for level plots of $ \cos x + \cos y $.
As a relevant aside I can also mention..
$$ \cos a \cos b = \cos c , \cosh\, a \cosh \,b = \cosh \,c $$ which are non-linear Non-Euclidean generalizations to linear Euclidean geometry also have their smaller degree approximation leading to the Pythagoras:
$$ a^2 + b^2 = c^2 $$
So by this token supposing you had asked " Why does $ \cosh x + \cosh y = \cosh(x+y) $ look like a hyperbola? ", the discussion would run, imho along similar lines.