Why is $T_n=Y_{n-1}+Y_n$ not a Markov chain? Let $(Y_n)_{n\in\mathbb{N}_0}$ be a sequence of iid random variables with an arbitrary distribution $q$ and existing expected value.
How can I prove that $T_n=Y_{n-1}+Y_n$ with $(T_n)_{n\in\mathbb{N}}$ is not a Markov chain?
How can I prove if $T_n$ is a Martingale?
 A: Still I maintain you haven't defined your sequence in $0$. What is $T_0$ ? So here, I assume $Y_n$ is define $\forall n \in \mathbb N_0$, and $T_n$ is $\forall n \in \mathbb N$.
Markov
$X_n$ is a Markov Chain 
$$\iff \mathbb E ( X_n | \mathcal F_{n-1} ) =  \mathbb E ( X_n | X_{n-1} ) $$
So, is $T_n$ a markov chain ? $T_n = Y_n + Y_{n-1} $
$$ \mathbb E ( T_n | \mathcal F_{n-1} ) = \mathbb E ( Y_n + Y_{n-1} | \mathcal F_{n-1} ) = \mathbb E ( Y_n ) + \mathbb E ( Y_{n-1} | Y_{n-1} )  
= \mathbb E ( T_n | Y_{n-1} ) $$
About those equalities, depending on your level of proficiency with probability theory, you might need some details, let me know about that. All comes down to basic properties, like the one with independance between the condition and the random variable inside the conditional expectation.
and $$ \mathbb E ( T_n | Y_{n-1} ) \neq   \mathbb E ( T_n | Y_{n-1} , Y_{n-2} ) = \mathbb E ( T_n | Y_{n-1} , Y_{n-2} ) = \mathbb E ( T_n | T_{n-1} ) $$
This is what's causing the problem according to me. You then have an example of a non markovian chain (first comment) given by Brian. However, you can easily modify your process to make it markovian with the proof I ve given you.
Also, check with the same steps what about the random walk $ T_n = T_{n-1} + Y_n $.

Martingale
$X_n$ is a Martingale
$$\iff \mathbb E ( X_n | \mathcal F_{n-1} ) =  X_{n-1} $$
Let's find a condition on the expectation of the $Y_n$.
$$ \mathbb E ( T_n | \mathcal F_{n-1} ) = \mathbb E ( Y_n + Y_{n-1} | \mathcal F_{n-1} ) = \mathbb E ( Y_n ) + \mathbb E ( Y_{n-1} | Y_{n-1} ) = \mathbb E ( Y_n ) + Y_{n-1} $$
can you conclude yourself ?
