Basic property of a tensor In Jost's Riemannian Geometry and Geometric Analysis (6th ed.) on page 142 there is the following remark concerning the torsion tensor.

Remark. It is not difficult to verify that [the torsion tensor] $T$ is indeed a tensor, i.e. that the value of $T(X,Y)(x)$ only depends on the values of $X$ and $Y$ at the point $x$.

Here, $X,Y\in\mathfrak{X}(M)$. 
Since a map $T:\mathfrak{X}(M)\times\mathfrak{X}(M)\to C^\infty(M)$ is induced by a tensor iff it is $C^\infty(M)$-linear and the torsion $T$ is indeed $C^\infty(M)$-linear, I know that the torsion $T$ is a tensor. However, I am not aware of the property of a tensor as stated above but supposing the property has something to do with the difference between tensor and tensor field. A clarification is needed: Where does the above property of a tensor come from?
 A: Before answering your question, let's clarify something else.
An $(r,s)$-tensor at $p \in M$ is an element of $\mathcal{T}^r_s(T_pM)$.  Thus, a tensor is an object defined at a single point; it is an element the total space of a vector bundle.
An $(r,s)$-tensor field is a section of the vector bundle $\mathcal{T}^r_s(TM)$.  Thus, a tensor field is a function which inputs points of a manifold and outputs tensors attached to those points; it is a section of a vector bundle.
In differential geometry, we rarely care about tensors defined at a single point as much as tensor fields. For this reason, many authors (including Jost) use the words interchangeably.  In this case, it would be more accurate to say that the torsion "tensor" is a tensor field.

Let $\pi \colon E \to M$ be a smooth vector bundle.
Let $F \colon \Gamma(E) \to C^\infty(M)$ be an $\mathbb{R}$-linear map.


*

*We say that $F$ acts locally iff for any two sections $\sigma_1, \sigma_2 \in \Gamma(E)$ which agree on an open set $U \subset M$ (i.e. $\sigma_1|_U = \sigma_2|_U$), we have $F(\sigma_1)|_U = F(\sigma_2)|_U$.

*We say that $F$ acts pointwise iff for any two sections $\sigma_1, \sigma_2 \in \Gamma(E)$ which agree at a point $p \in M$ (i.e. $\sigma_1(p) = \sigma_2(p)$), we have $F(\sigma_1)(p) = F(\sigma_2)(p)$.

Fact: An $\mathbb{R}$-linear map $F\colon \Gamma(E) \to C^\infty(M)$ acts pointwise iff $F$ is $C^\infty(M)$-linear.

Proof sketch: $(\Longrightarrow)$ Suppose $F \colon \Gamma(E) \to C^\infty(M)$ acts pointwise. Fix $p \in M$ and $f \in C^\infty(M)$. Note that the sections $\sigma_1 := f\sigma$ and $\sigma_2 := f(p)\sigma$ have $\sigma_1(p) = f(p)\sigma(p) = \sigma_2(p)$, so that
$$F(f\sigma)(p) = F(f(p)\sigma)(p) = f(p)(F\sigma)(p).$$
That is, $F(f\sigma) = f(F\sigma)$.
$(\Longleftarrow)$ Suppose $F$ is $C^\infty(M)$-linear.  Let $U \subset M$ be open, $p \in U$.  Suppose $\sigma|_U = 0$.  Let $\varphi \in C^\infty(M)$ be a bump function with $\text{supp}(\varphi) \subset U$ and $\varphi(p) = 1$. Note that $\varphi \sigma = 0$ (on $M$).  By $C^\infty(M)$-linearity, we have $\varphi F(\sigma) = F(\varphi \sigma) = 0$.  Thus, $F(\sigma)(p) = \varphi(p)F(\sigma)(p) = 0$, so $F(\sigma)|_U = 0$.  This shows that $F$ acts locally.  By using another bump function argument, one can show (I omit the details) that $F$ acts pointwise.
Reference: "Introduction to Smooth Manifolds" by John Lee
