Why does spacetime exhibit a different geometry to space? In his lectures, Feynman compares the transformations for rotations of coordinate space - $x'= x\cos\alpha + y\sin\alpha$, $y' = y\cos\alpha - x\sin\alpha$ and z' = z - with the Lorentz transformations, both being effective 'rotations' in space and spacetime respectively. He then writes that the difference in sign between the two sets of transformations and the fact that the rotation in space is written in terms of $\cos(\alpha)$ and $\sin(\alpha)$, rather than algebraic quantities, leads to a difference in geometry. I am struggling to get an intuition for why this is the case, though it makes sense mathematically. I would appreciate an intuition for this.
 A: 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).)

A: A geometric visualization is sectioning a cone by a plane perpendicular to its axis in the case of a 2D Euclidian space $x,y$. The result is a cicle. The radius of the circle obviously is conserved by rotations.
In the case of a 2D Minkowskian spacetime $t,x$, the plane is parallel to the axis, and the result are 2 hyperbolas for a double cone. Any line segment,
joining the symmetry center of this plane figure to a point in one of the hyperbolas, conserves the quantity $t^2 - x^2$, which is a consequence of Lorentz transformation. This transformation can be viewed as the rotation of that line. The tangent of the angle with the axis $t$ is the speed of the moving frame.
A: There are good answers already giving a mathematical perspective. If these don't help your intuition, a high level explanation is that time and space are related but not the same. Rotations mixing time and space cannot be the same as purely spatial rotations; otherwise we could turn around in time and violate causality. The geometry of spacetime reflects this difference between space and time.
