Constructing Turing machines
It is simpler than what you think. You don't even need a Turing machine that writes to the tape. 

Here is a Turing machine whose final state would be $YES$ if the string (in tape) contains an even number of 1's, and in $NO$ if the string contains an odd number of 1's, assuming the string marks its end with a #, and it starts from left to right.

Let $ M = (Q, Γ, Σ, δ, s, B, F)$ be a Turing Machine with the definition 

$$Q = \{read_0,read_1,YES,NO\}$$

$$Γ = \{1,0,\#\}$$

$$Σ = \{\}$$

$$F = \{YES,NO\}$$

$$s = read_0$$

$$B=\#$$

$δ$ is given by $δ_{q\ t} = ( move \,\ next\ stage)$ where $t$ is the tape symbol that it reads, and $q$ represents each state:

$$δ_{read_0\ 0} = (R,read_0)$$ 
$$δ_{read_0\ 1} = (R,read_1)$$ 
$$δ_{read_0\ \#} = (0,YES)$$ 
$$δ_{read_1\ 0} = (R,read_1)$$
$$δ_{read_1\ 1} = (R,read_0)$$ 
$$δ_{read_1\ \#} = (0,NO)$$



Basically what it does is that if it see $1$ it alters between $read_0$ and $read_1$ states. And if it is in $read_0$ state and sees $\#$, which means the string ended, it would go to the final state, $YES$. If it is in $read_1$ and sees $\#$ it would go to $NO$.

So if the string contains three $1$s, the states change in this way $$read_0 \rightarrow read_1 \rightarrow read_0 \rightarrow read_1 \rightarrow NO$$ implying that it doesn't contain an even number of 1s.