You didn't say which CF4 manual you are looking at. Looking at the MCF54455 Reference manual and searching for "Harvard" gets:
3.1.1 Overview mentions "Harvard".
4.1.2 Features says "(Harvard TLBs)" twice.
Table 4-9. MMUDR Field Descriptions says "If a Harvard TLB implementation is used" and "If separate ITLB and DTLBs are used,"
4.3.8 MMU Implementation says "The MMU implements a 64-entry full-associative Harvard TLB architecture".
4.3.8.3 TLB Locked Entries says "Figure 4-12 is a ColdFire MMU Harvard TLB block diagram".
Figure 4-12. Version 4 ColdFire MMU Harvard TLB.
I would guess the MMUDR description is describing a GENERIC MMU Module which could be reused in different CPUs, some with and others without q Harvard architecture.
In this case the CPU is certainly "a Harvard TLB implementation".
If you've ever read any ARM documentation you'll find it is all like this. The chip manual states something like "Contains a Cortex A8 core, go read the ARM documentation". Since the cores come with a lot of chip-manufacturer-selectable options (for optional features and different cache sizes and so on) the "generic" documentation doesn't detail things like that. Sometimes the "bus interconnect" from the core to the rest of the chip is selectable and separately documented.
> My goal here is to be able to both read and execute a page of instruction memory and use *only* one entry in the table, rather than two TLB entries
I can see why you would want to do this, but I don't think it is possible.
You might find some example code here if you can work through the twisty maze that defines what gets built linked and connected:
Linux/arch/m68k/mm/mcfmmu.c - Linux Cross Reference - Free Electrons
Tom