Let $Z = -Y$ so that $X-Y = X+Z$. Then, $$\operatorname{var}(Z)= (-1)^2\operatorname{var}(Y) = \operatorname{var}(Y)$$ and
$$\operatorname{cov}(X,Z) = \operatorname{cov}(X,-Y) = -\operatorname{cov}(X,Y)$$
so that
\begin{align}
\operatorname{var}(X-Y) &= \operatorname{var}(X+Z)\\
&= \operatorname{var}(X)+\operatorname{var}(Z) + 2\operatorname{cov}(X,Z)\\
&= \operatorname{var}(X)+\operatorname{var}(Y) - 2\operatorname{cov}(X,Y)
\end{align}
In short,
\begin{align}
\operatorname{var}(X+Y) &= \operatorname{var}(X) + \operatorname{var}(Y) + 2\operatorname{cov}(X,Y)\\
\operatorname{var}(X-Y) &= \operatorname{var}(X) + \operatorname{var}(Y) - 2\operatorname{cov}(X,Y)
\end{align}
really are the same formula, and which of the two variances is larger depends entirely on whether $\operatorname{cov}(X,Y)$ is positive or negative. In the case when $\operatorname{cov}(X,Y)$ equals $0$ (i.e. $X$ and $Y$ are uncorrelated random variables), $\operatorname{var}(X+Y)$ equals $\operatorname{var}(X) + \operatorname{var}(Y)$ as you have already noted.
As to intuition regarding why $\operatorname{var}(X+Y) > \operatorname{var}(X-Y)$ when $\operatorname{cov}(X,Y) > 0$, a geometric viewpoint might help. $X$ and $Y$ can be regarded as vectors of lengths $\sigma_X = \sqrt{\operatorname{var}(X)}$ and $\sigma_X = \sqrt{\operatorname{var}(Y)}$ that are pointed roughly in the same direction when $\operatorname{cov}(X,Y) > 0$ and in roughly opposite direction when $\operatorname{cov}(X,Y) < 0$. Now, one would expect intuitively that the
(vector) sum of two vectors pointing in roughly the same direction is a longer vector than the (vector) difference of the two vectors, no? A crudely drawn picture of the difference vector and the sum vector is in the diagram below.
Note that in the left-hand figure above, the vectors $X$ and $Y$ are shown as being at an acute angle $\theta$ and thus are "pointed in roughly the same direction". In fact, the correlation coefficient $\rho$ of (random variables) $X$ and $Y$ is $\cos(\theta) > 0$. Now, in a triangle with vertices $A, B, C$, and opposite sides of lengths $a, b$ and $c$ respectively, we have the following result familiar from elementary geometry/trigonometry
$$c^2 = a^2 + b^2 - 2ab\cos(\angle C)$$ which is equivalent to
\begin{align}
\sigma_{X-Y}^2 &= \sigma_X^2 + \sigma_Y^2 - 2\sigma_X\sigma_Y\cos(\theta)\\
\operatorname{var}(X-Y) &= \operatorname{var}(X) + \operatorname{var}(Y)
-2 \rho \sigma_X\sigma_Y\\
\operatorname{var}(X-Y) &= \operatorname{var}(X) + \operatorname{var}(Y)
-2 \operatorname{cov}(X,Y)
\end{align}
On the other hand, as shown in the right-hand figure above, for vector sums,
the included angle $\pi-\theta$ is obtuse, and so
$\cos(\pi-\theta) = -\cos(\theta) < 0$. So, we get that
\begin{align}
\sigma_{X+Y}^2 &= \sigma_X^2 + \sigma_Y^2 + 2\sigma_X\sigma_Y\cos(\theta)\\
\operatorname{var}(X+Y) &= \operatorname{var}(X) + \operatorname{var}(Y)
+2 \operatorname{cov}(X,Y),
\end{align}
that is, $\operatorname{var}(X+Y) > \operatorname{var}(X-Y)$ for positively correlated random variables.
