You're on the right general track, but there are some small algebraic mistakes.
First, note that the number of odd digits (or the number of even digits) is $x$, not $x/2$, since the total number of digits is $2x$; basically, the division by 2 that you're using in your current (incorrect) formula for $t_{even}$ has already been 'baked in' for you. Likewise, the number of even digits is $x+1$ : $0, 2, \ldots, 2x$ (there's a sneaky off-by-one problem lurking here, one that I've made a couple of times already in this answer and its comments!)
Second and more critically, there are errors in both your terms because you're missing a factor of $x$. In your $t_{odd}$ term, you're right about the ways it can break down but forgot that there can be $x$ different 'next digits'; you make the same mistake in your $t_{even}$ term, one layer removed. Try working through your tentative recurrences in the case $n=3, x=2$ (so there are 5 digits) to see what I mean.
Once you take these factors into account, you'll have an expression for $t_{odd}(n)$ in terms of a constant factor times $(t_{odd}(n-1)+t_{even}(n-1))$, and likewise an expression for $t_{even}(n)$ in terms of a constant factor times $(t_{odd}(n-2)+t_{even}(n-2))$; now just add them, set $u(n)=t_{odd}(n)+t_{even}(n)$, and note that this lets you write $u(n)$ in terms of $u(n-1)$ and $u(n-2)$.
Note that you can get at the same result somewhat more directly; let $u(n)$ be the number of such strings with $n$ digits (i.e., $u(n)=t_{odd}(n)+t_{even}(n)$.) Then as you already noticed, if $u(n)$ begins with an odd digit, the rest of the sequence can be arbitrary, while if $u(n)$ begins with an even digit, the next digit must be odd but the rest can be arbitrary; in other words, $u(n) = \{\#$of ways the first digit can be odd$\}\cdot u(n-1)+\{\#$of ways the first digit can be even$\}\cdot\{\#$of ways the second digit can be odd$\}\cdot u(n-2)$, and all of the bracketed expressions should be easy.
