$\let\s=\sigma \let\pd=\partial \let\dn=\downarrow \let\up=\uparrow
\let\dag=\dagger
\def\hp{\hat p} \def\hs{\hat s} \def\hx{\hat x} \def\hy{\hat y}
\def\hz{\hat z} \def\hI{{\hat I}} \def\hL{{\hat L}}
\def\hP{{\skew{1}\hat P}} \def\hT{{\skew{1}\hat T}}
\def\ket#1{|#1\rangle} \def\bra#1{\langle#1|}
\def\braket#1#2{\langle#1|#2\rangle}
\def\mxelm#1#2#3{\bra#1\,#2\,\ket#3}
\def\half{{\textstyle{1 \over 2}}}
\def\kd{\ket{\!\dn\,}} \def\ku{\ket{\!\up\,}}$
The answer is in keeping a clear distinction between abstract
operators and their representations. Which implies a parallel
distinction between abstract Hilbert spaces and their representations.
The simple equation you wrote
$$T = K\,U \tag1$$
doesn't and can't hold for abstract operators. In an abstract Hilbert
space conjugation is meaningless (or, if you like better, can be
defined in infinite different ways, so it's useless). Instead eq. (1) is
meaningful if you're referring to a well defined representation, and
the $U$ will be different in different representations of the same
abstract space.
I'll show you some examples. First, a particle moving along a
straight line. All observables are definable starting from position
$\hx$ and momentum $\hp$. So to define time reversal it suffices to
give its action on $\hx$ and $\hp$:
$$\hT\,\hx\,\hT^\dag = \hx \qquad \hT\,\hp\,\hT^\dag = -\hp.\tag2$$
Note that $\hT$ preserves commutation relation
$$[\hx,\hp] = i\,\hbar\,\hI.$$
In fact applying time reversal to LHS
$$\hT\,[\hx,\hp]\, \hT^\dag = [\hx,-\hp] = -i\,\hbar\,\hI$$
and applying it to RHS
$$\hT\,(i\,\hbar\,\hI)\,\hT^\dag = -i\,\hbar\,\hI$$
because $\hT\,i\,\hT^\dag=-i$ (antiunitarity). Note that for a
antiunitary operator, as for a unitary one, $\hT^\dag=\hT^{-1}$.
The $\hat{}$ over operator symbols is there to remind you that these
are abstract operators. I haven't yet defined any representation.
For a particle on a line the most common representation is
Schrödinger's:
$$\hx \mapsto x \qquad \hp \mapsto -i\,\hbar{\pd \over \pd x}.$$
Then you can see that in order to satisfy (2) we just have to put
$$\hT \mapsto K$$
where $K$ means complex conjugation of wavefunctions and of operator
representations. In fact $Kx=x^*=x$, whereas momentum changes sign
because of the "$i$" (BTW $K^\dag=K$).
Another useful representation for the Hilbert space of this physical
system is momentum representation. It is defined by $$\hx \mapsto
i\,\hbar{\pd \over \pd x} \qquad \hp \mapsto p.$$ If you tried to use
again $\hT\mapsto K$ you'd get a wrong result: $\hp$ unchanged and
$\hx$ reversed! The right solution is
$$\hT \mapsto K\,P$$
where $P$ is the space inversion operator, which changes sign both
to $x$ and to $p$:
$$\hP\,\hx\,\hP^\dag = - \hx \qquad \hP\,\hp\,\hP^\dag = -\hp.$$
So you see that eq. (1) holds in both representations, but with
different $U$'s: $U=I$ in Schr. rep, $U=P$ in momentum rep.
A second example is a spin-1/2 particle. If we are only concerned with
spin states, the only observables are spin components. There are three
linearly independent, say $\hs_x$, $\hs_y$, $\hs_z$. The system's
Hilbert space actually is just a 2D complex vector space. In it a
basis is customarily assumed consisting of eigenvectors of $\hs_z$:
$\ku$, $\kd$
$$\hs_z \ku = \half\,\hbar\,\ku \qquad
\hs_z \kd = -\half\,\hbar\,\kd.$$
Note that this doesn't completely define the basis, as arbitrary phase
factors are still allowed for both vectors. In order to maximally
define the basis the action of another observable is needed, e.g.
$\hs_x$. The usual definition is
$$\hs_x \ku = \half\,\hbar\,\kd \qquad
\hs_x \kd = \half\,\hbar\,\ku$$
Once a basis is fixed, a representation of spin observables is also
defined. It is
$$\hs_x \mapsto \half\,\hbar\,\s_x \qquad
\hs_y \mapsto \half\,\hbar\,\s_y \qquad
\hs_z \mapsto \half\,\hbar\,\s_z$$
where
$$\s_x = \pmatrix{0&1\cr1&0\cr} \qquad
\s_y = \pmatrix{0&-i\cr i&0\cr} \qquad
\s_z = \pmatrix{1&0\cr0&-1\cr}$$
are the usual Pauli matrices.
Now the action of $\hT$ on spin observables is to be defined. If we
want to ensure time reversal invariance say for an atom, we must take
for $\hs_x$, $\hs_y$, $\hs_z$, the same transformation law holding
for orbital angular momentum. This is already fixed by transformation
laws of $\hx$, $\hy$, $\hz$, $\hp_x$, $\hp_y$, $\hp_z$: the former are
invariant, the latter change sign. Then $\hL_z$, being defined as
$\hx\,\hp_y-\hy\,\hp_x$, changes sign too, and the same happens for
the other components. We conclude that all spin components change
sign.
Once again we see that the simple choice $\hT\mapsto K$ doesn't work:
it's OK for $\hs_y$ but not for the other two. Remembering that
1/2-spin components anticommute with one another the solution is
immediate:
$$\hT \mapsto K \s_y.$$
I leave for you to check this is the right answer.
