Edited. Here is what is going on. If the first determinant is zero, then the inequality follows immediately, But if the first determinant is not zero, then it need not hold.
Here is exactly what happens, and how to get as many counterexamples as you please.
Take any matrix whose determinant is not zero: $\left(\begin{array}{cc} a&c\\ b&d \end{array}\right),$ and let $\left|\det\left(\begin{array}{cc} a & c\\ b & d\end{array}\right)\right| = D.$
If $D\neq 0$, that means that the columns, $(a,b)$ and $(c,d)$, form a basis for $\mathbb{R}^2$. Hence, every vector $(e,f)$ can be written uniquely as $(e,f)=\alpha(a,b) + \beta(c,d)$ for some choice of scalars $\alpha$ and $\beta$. Thus, by the properties of the determinant, we have: \begin{align*} \left|\det\left(\begin{array}{cc} a & e\\ b & f \end{array}\right)\right| &= \left|\det\left(\begin{array}{cc} a & \alpha a+\beta c\\ b & \alpha b + \beta d \end{array}\right)\right| \\ &= \left|\det\left(\begin{array}{cc} a&\alpha a\\ b &\alpha b \end{array}\right) + \det\left(\begin{array}{cc} a & \beta c\\ b & \beta d \end{array}\right)\right|\\ &= \left| 0 + \beta\det\left(\begin{array}{cc} a & c\\ b & d \end{array}\right)\right|\\ &=|\beta|D. \end{align*} Symmetrically, you will have that $\left|\det\left(\begin{array}{cc} e & c\\\ f & d \end{array}\right)\right| = |\alpha|D.$ So you will have $ \left|\det\left(\begin{array}{cc} a & e\\\ b & f \end{array}\right)\right| + \left|\det\left(\begin{array}{cc} e & c\\\ f & d \end{array}\right)\right| = |\alpha|D + |\beta|D = (|\alpha|+|\beta|)D,$ where $\alpha$ and $\beta$ are the unique scalars such that $(e,f) = \alpha(a,b) + \beta(c,d)$.
So the inequality you want is equivalent to $D\leq (|\alpha|+|\beta|)D,$ which holds if and only if $|\alpha|+|\beta|\geq 1$, but there are plenty of choices that will make $|\alpha|+|\beta| \lt 1$, in which case your inequality cannot hold.
My first example, now deleted, took $(a,b)=(1,0)$, $(c,d) = (0,1)$, $\alpha=e$ and $\beta=0$. The second example (also now deleted in favor of this exposition) took $\alpha=\beta=e$. As long as neither of $\alpha$ and $\beta$ are zero, you'll get nondegenerate parallelograms, but picking them small in absolute value will ensure the inequality cannot hold.