You are mistaken on the definition of $T$. Since $T$ is an application from $P(R)$ into $P(R)$ (I assume that $P(R)$ stands for the vector space of polynomials with real coefficients), the input is a polynomial $p$ and the output is also a polynomial function $T(p)$, usually written as $Tp$.
What you have to show in order to prove that $T$ is linear is:
1. for any $p$ and $q$ in $P(R)$, $T(p+q)=T(p)+T(q)$.
2. for any $p$in $P(R)$ and any real number $c$, $T(cp)=cT(p)$.
Then... what is that $x$ in the definition of $T$? Since $f=T(p)$ is itself a function, it is defined by expliciting its evaluation $f(x)$ at every real number $x$.
To sum up, $T$ is an application (= a function) that transforms a polynomial function $p$ into a new ploynomial application $f=T(p)$.
Let's prove the point $1$ above (point $2$ is proved similarly). Let $p$ and $q$ be two polynomials. Then, we need to check the equality of functions $T(p+q)=T(p)+T(q)$. For, consider a real number $x$. We have
$$T(p+q)(x)=x^2(p+q)(x)=x^2(p(x)+q(x))=x^2p(x)+x^2q(x) = T(p)(x) + T(q)(x)$$
that is $T(p+q)(x) = (T(p)+T(q))(x)$. The claim $T(p+q)=T(p)+T(q)$ is proved! Make sure you understand each step of the computation above, this kind of game with definitions/formulas is important for abstract algebra.
Note: the notation $T(p)(x)$ means that the function $T(p)$ is evaluated at $x$. If you let $f=T(p)$, then it is just $f(x)$.
Note 2: in general, polynomials are not linear applications, so we don't have $p(x+y)=p(x)+p(y)$ and $p(cx)=cp(x)$.
Note 3: for simplicity, I identified polynomials and polynomial functions, though they are not exactly the same. It is fine because we are considering real coefficients here.