The difference of the geometries is due to certain sign differences and the use of hyperbolic-trig functions vs circular-trig functions --- not the use of algebraic variables vs trig variables.
Note that rotations can written in terms of slope as
$$x'=x\frac{1}{\sqrt{1+m^2}}+y\frac{m}{\sqrt{1+m^2}}$$
$$y'=x\frac{-m}{\sqrt{1+m^2}}+y\frac{1}{\sqrt{1+m^2}}$$
where $m=\tan\alpha$,
and Lorentz boosts can be written trigonometrically as
$$t'=t\cosh\theta+\frac{y}{c}\sinh\theta$$
$$\frac{y}{c}'=t\sinh\theta+\frac{y}{c}\cosh\theta$$
where $\frac{v}{c}=\tanh\theta$.
Both are examples of affine Cayley-Klein geometry (see also https://en.wikipedia.org/wiki/Cayley%E2%80%93Klein_metric#Special_relativity ) (which involves projective geometry).
UPDATE:
I think it is enlightening to point out that this "difference of geometry" already occurs with Galilean relativity. (Special relativity made us more aware of this situation... and we see this when we look back to Galilean physics.)
It is known (but not well known) that the position-vs-time graph (regarded as a Galilean spacetime) does not have a Euclidean geometry. In general, two inertial worldline segments with equal elapsed times do not necessarily have equal Euclidean lengths. [Take one inertial worldline to be at rest, and the other at traveling at 3 m/s. Suppose they met at event O. Stop the diagram 1-second after event O. Although their elapsed times are equal to 1, the drawn-length of their worldline segments are unequal.]
(The position-vs-time graph also has a different metric and, thus, a different "circle" and a different notion of perpendicularity from that of Euclidean geometry.)
Thus, the position-vs-time graph (from PHY 101) has a non-Euclidean geometry (with zero curvature).
Aspects of this can be seen in my visualization
https://www.desmos.com/calculator/kv8szi3ic8
(Tune the E-slider from 1(Minkowski) to -1(Euclidean), and observe the case E=0 (Galilean).)
The generalization to four-dimensional Lorentz transformations
is now quite natural:
$$J^{\mu \nu} = -i(x^{\mu}\partial^\nu - x^{\nu}\partial^\mu).$$
You need to be very careful about upper and lower indices here.
Spacetime position is defined as the 4-vector
$$x^\mu=(ct,x,y,z). \tag{1}$$
These are contravariant components, i.e. with an upper index.
Likewise the gradient is defined as the 4-vector
$$\partial_\mu=\frac{\partial}{\partial x^\mu}=
\left(\frac{\partial}{c\ \partial t},
\frac{\partial}{\partial x},
\frac{\partial}{\partial y},
\frac{\partial}{\partial z}\right) \tag{2}$$
Notice, these are its covariant components (with lower index).
But the contravariant components (with upper index)
of the gradient are slightly different:
$$\partial^\mu==
\left(\frac{\partial}{c\ \partial t},
-\frac{\partial}{\partial x},
-\frac{\partial}{\partial y},
-\frac{\partial}{\partial z}\right) \tag{3}$$
This follows from raising and lowering indices
$A^\mu=\eta^{\mu\nu}A_\nu$ with the Minkowski metric.
I'm using the $(+,-,-,-)$ metric sign convention here.
If we wanted to represent rotations/angular momentum in 4 dimensional
space, i.e. $SO(4)$, then this generalization would seem appropriate.
You are right. The Lorentz group is not $SO(4)$, but it is $SO(1,3)$.
But boosts are similar to but not identical with a 4d rotation.
Why does the generalization seem to look like space and time are
treated exactly equally?
Take for example the $^{01}$ component of Peskin's formula
$$J^{\mu\nu} = -i(x^{\mu}\partial^\nu - x^{\nu}\partial^\mu).$$
Using (1) and (3) we get the boost in $x$ direction
$$J^{01} = -i(x^0\partial^1 - x^1\partial^0)
=-i\left(-ct\frac{\partial}{\partial x}
- x\frac{\partial}{c\ \partial t}\right).$$
Notice that both terms got the same sign, as it should be
for a Lorentz boost.
This is slightly different from a Lorentz rotation, for example
the rotation around the $z$ axis
$$J^{12} = -i(x^1\partial^2 - x^2\partial^1)
=-i\left(-x\frac{\partial}{\partial y}
+ y\frac{\partial}{\partial x}\right)$$
where both terms have opposite signs.
So when only looking at the formula for $J^{\mu\nu}$ it may
seem, time and space dimensions are treated exactly equally.
But actually there is a subtle difference between time and space
dimensions because of the minus signs in formula (3) for $\partial^\mu$,
which in turn come from the the minus signs in the Minkowski
metric $\eta^{\mu\nu}$.
Best Answer
The relevant generalized notion of "rotation" is that a rotation is a transformation that fixes one point and preserves all distances. In euclidean space, this means that if you have two points with x-coordinates differing by $\Delta x$ and y-coordinates ďiffering by $\Delta y$, then the value of $\Delta x^2+\Delta y^2$ is unaffected by a rotation. In Minkowski space it means that $\Delta x^2-\Delta t^2$ is unaffected.