= Solution
For $X\subseteq\mathbb A^n$ and $P\in X$, the <Zariski tangent space> is
$$
T_PX=\{v\in k^n:df_P(v)=0\text{ for every }f\in I(X)\}.
$$
The dimension can be defined as $\min_{P\in X}\dim T_PX$. Equivalently, the <krull dimension of an affine variety> is the supremum of lengths of strict chains of irreducible closed subsets, or the Krull dimension of $k[X]$. A point is singular when $\dim T_PX>\dim X$.
For
$$
X=Z(x_1^2-x_2^2,x_3^2-x_4^2),
$$
the differentials give the <tangent spaces of a product of two nodal line pairs>
$$
\boxed{T_PX=\{v:x_1v_1-x_2v_2=0,\quad
x_3v_3-x_4v_4=0\}}.
$$
The variety is a union of four two-dimensional linear spaces. If both pairs $(x_1,x_2)$ and $(x_3,x_4)$ are nonzero, the two displayed equations are independent and $\dim T_PX=2$. If exactly one pair is zero, the dimension is three; at the origin it is four. Thus the singular locus is the union of the loci where either coordinate pair vanishes.
For $Y$, rank at most one is equivalent to vanishing of all $2\times2$ minors:
$$
y_0y_3-y_1y_2=0,\qquad
y_0y_4-y_2^2=0,\qquad
y_1y_4-y_2y_3=0.
$$
These homogeneous equations prove that $Y$ is projectively Zariski closed.
The following affine charts show both its dimension and smoothness. On $y_0=1$,
$$
y_3=y_1y_2,\qquad y_4=y_2^2,
$$
so $(y_1,y_2)$ are free. On $y_1=1$,
$$
y_2=y_0y_3,\qquad y_4=y_0y_3^2,
$$
so $(y_0,y_3)$ are free. On $y_3=1$,
$$
y_2=y_1y_4,\qquad y_0=y_1^2y_4,
$$
and on $y_4=1$,
$$
y_1=y_2y_3,\qquad y_0=y_2^2.
$$
These four charts cover $Y$, since the point with only $y_2$ nonzero violates the second minor. Every chart is isomorphic to $\mathbb A^2$. Hence the <smooth projective surface from overlapping rank-one coordinates> satisfies
$$
\boxed{\dim Y=2}
$$
and is nonsingular everywhere.
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