3 improved formatting, fixed typos

Here is the answer to your specific question (and, in fact, something a little more general):

The hypersurface defined by $f=x_{ij}-x_{kl}$ (assuming that $i$, $j$, $k$, and $l$ are distinct) is smooth if $n\le 5$ and singular for $n>5$. The singular locus consists of the $2$-planes that lie in the codimension $4$ subspace that consists of $0$ plus the vectors $v$ such that pairing $v$ with any linearly independent vector generates a $2$-plane that still lies in the hypersurface.

The way to think about this more geometrically is this: Take any nonzero $2$-form $\Omega$ on $\mathbb{C}^n$. Consider the set $Z_\Omega\subset Gr(2,n)$ that consists of the $2$-planes on which $\Omega$ vanishes. This is a hypersurface in $Gr(2,n)$ of exactly the kind you are considering in your specific problem. Let $N\subset\mathbb{C}^n$ be the null space of $\Omega$, i.e., the subspace consisting of the vectors $v\in\mathbb{C}^n$ such that $\Omega(v,w) = 0$ for all $w\in\mathbb{C}^n$. Then a $2$-plane $V$ is a singular point of $Z_\Omega$ if and only if $V$ is a subspace of $N$.

In your particular case, $\Omega = dz^i\wedge dz^j - dz^k\wedge dz^l$, so $N$ is the subspace defined by $z^i = z^j = z^k = z^l = 0$.

In the more general case, the codimension of $N$ is twice the largest integer $\rho$ such that $\Omega^\rho\not=0$.

These claims are easily verified in a chart on $Gr(2,n)$.

Addendum: I forgot to answer the question about whether the hypersurface $Z_\Omega$ is irreducible. It is. The reason is that the smooth locus is connected, as follows without too much difficulty from the above description.

Also, you asked whether the intersection of a number of these 'linear' hypersurfaces is always connected. The answer to this is 'no'. For a simple example, take the locus defined by $x_{13}=x_{14}=x_{23}=x_{24}=0$ in $Gr(2,4)$. This consists of two points, a pair of 2-planes in general position in $\mathbb{C}^4$.

In general, deciding whether such an intersection is smooth or connected (or even nonempty) is not easy. In the theory of exterior differential systems, there is a criterion that is sufficient for a point in $Gr(2,n)$ to be smooth that is often useful, even though it is far from necessary. I'll describe it briefly here, but you can get more information in our book Exterior Differential Systems.

For notational simplicity, let me denote $\mathbb{C}^n$ by $W$. Let $\Sigma\subset \Lambda^2(W^\ast)$ be a linear subspace, and let $Z_\Sigma\subset Gr(2,W)$ denote the set of $2$-planes $V\subset W$ such that all of the elements of $\Sigma$ vanish on $V$. I want to define the locus $Z^o_\Sigma\subset Z_\Sigma$ of ordinary elements, and these will be smooth points of $Z_\Sigma$. To do this, for each $w\in W$, consider the vector space $H(w)\subset W$ consisting of all the vectors $v\in W$ such that $\Omega(v,w)=0$ for all $\Omega\in\Sigma$. Say that $w$ is $\Sigma$-regular if the dimension of $w$ H(w)$is minimal among all$w\in W$. (The$\Sigma$-regular elements in$W$form a Zariski-open subset of$W$.) Say that$V\in Z_\Sigma$is$\Sigma$-ordinary if$V$contains a$\Sigma$-regular vector. Then it It is not hard to prove show that the set$Z^o_\Sigma\subset Z_\Sigma$consisting of$\Sigma$-ordinary elements consists of smooth points of$Z_\Sigma$, and its closure is an irreducible component of$Z_\Sigma$. However, the set of ordinary points$Z^o_\Sigma$might well be empty, even though$Z_\Sigma$is nonempty (and it even could even be smooth). For the purposes of exterior differential systems, though, these turn$Z^o_\Sigma$turns out to be the most interesting points part of$Z_\Sigma$(when they exist)it is nonempty). 2 added information and answers Addendum: I forgot to answer the question about whether the hypersurface$Z_\Omega$is irreducible. It is. The reason is that the smooth locus is connected, as follows without too much difficulty from the above description. Also, you asked whether the intersection of a number of these 'linear' hypersurfaces is always connected. The answer to this is 'no'. For a simple example, take the locus defined by$x_{13}=x_{14}=x_{23}=x_{24}=0$in$Gr(2,4)$. This consists of two points, a pair of 2-planes in general position in$\mathbb{C}^4$. In general, deciding whether such an intersection is smooth or connected (or even nonempty) is not easy. In the theory of exterior differential systems, there is a criterion that is sufficient for a point in$Gr(2,n)$to be smooth that is often useful, even though it is far from necessary. I'll describe it briefly here, but you can get more information in our book Exterior Differential Systems. For notational simplicity, let me denote$\mathbb{C}^n$by$W$. Let$\Sigma\subset \Lambda^2(W^\ast)$be a linear subspace, and let$Z_\Sigma\subset Gr(2,W)$denote the set of$2$-planes$V\subset W$such that all of the elements of$\Sigma$vanish on$V$. I want to define the locus$Z^o_\Sigma\subset Z_\Sigma$of ordinary elements, and these will be smooth points of$Z_\Sigma$. To do this, for each$w\in W$, consider the vector space$H(w)\subset W$consisting of all the vectors$v\in W$such that$\Omega(v,w)=0$for all$\Omega\in\Sigma$. Say that$w$is$\Sigma$-regular if the dimension of$w$is minimal among all$w\in W$. Say that$V\in Z_\Sigma$is$\Sigma$-ordinary if$V$contains a$\Sigma$-regular vector. Then it is not hard to prove that the set$Z^o_\Sigma\subset Z_\Sigma$consisting of$\Sigma$-ordinary elements consists of smooth points of$Z_\Sigma$, and its closure is an irreducible component of$Z_\Sigma$. However, the set of ordinary points might well be empty, even though$Z_\Sigma$is nonempty (and it even could be smooth). For the purposes of exterior differential systems, though, these turn out to be the most interesting points of$Z_\Sigma$(when they exist). 1 Here is the answer to your specific question (and, in fact, something a little more general): The hypersurface defined by$f=x_{ij}-x_{kl}$(assuming that$i$,$j$,$k$, and$l$are distinct) is smooth if$n\le 5$and singular for$n>5$. The singular locus consists of the$2$-planes that lie in the codimension$4$subspace that consists of$0$plus the vectors$v$such that pairing$v$with any linearly independent vector generates a$2$-plane that still lies in the hypersurface. The way to think about this more geometrically is this: Take any nonzero$2$-form$\Omega$on$\mathbb{C}^n$. Consider the set$Z_\Omega\subset Gr(2,n)$that consists of the$2$-planes on which$\Omega$vanishes. This is a hypersurface in$Gr(2,n)$of exactly the kind you are considering in your specific problem. Let$N\subset\mathbb{C}^n$be the null space of$\Omega$, i.e., the subspace consisting of the vectors$v\in\mathbb{C}^n$such that$\Omega(v,w) = 0$for all$w\in\mathbb{C}^n$. Then a$2$-plane$V$is a singular point of$Z_\Omega$if and only if$V$is a subspace of$N$. In your particular case,$\Omega = dz^i\wedge dz^j - dz^k\wedge dz^l$, so$N$is the subspace defined by$z^i = z^j = z^k = z^l = 0$. In the more general case, the codimension of$N$is twice the largest integer$\rho$such that$\Omega^\rho\not=0$. These claims are easily verified in a chart on$Gr(2,n)\$.