In the NTRU cryptosystem we are dealing with convolution polynomial rings and we compute
$f(x)= T(d+1,d)$ and $g(x)= T(d,d)$ but when calculating their inverse in $R_q=(Z/qZ[x] / (x^N-1))$ and $R_p$ why can we only use $f(x)$ and not $g(x)$?
Definition :T(d1,d2) = a(x) : a(x) has d1 coefficients equal to 1 and d2 coefficients equal to -1
I am really guessing as to what the question really is, and my guess is that the OP wants to know why $f(x)$ has an inverse in the ring $R_q$ but $g(x)$ does not.
If $g(x)$ has the same number of +1 and -1s as its coefficients this means that $(x-1)\mid g(x)$. As $x-1$ is a factor of $x^N-1$ this implies that $g(x)$ can never be an invertible element in $R_q$. For all elements $b(x)$, we have that the remainder of $g(x)b(x)$ modulo $x^N-1$ will be divisible by $x-1$ because both $g(x)$ and $x^N-1$ are.
The same argument does not apply to $f(x)$, because $f(x)\equiv1\pmod{x-1}$. Mind you this does not guarantee that $f(x)$ will have an inverse. It is still possible that $f(x)$ and $x^N-1$ have another common factor in the ring $\Bbb{Z}_q[x]$. IIRC the NTRU choices for $q$ and $N$ are designed with the view (among other things) in mind that $x^N-1$ has relatively few non-trivial factors (and these of a relatively high degree). This has the effect that with a ridiculously high probability the polynomial $f(x)$ has an inverse, if it does not have the trivial factor $x-1$.