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I know that $T_{(p,q)} M \times N$ can be identified with $T_pM \oplus T_q N$, but given $v \in T_{(p,q)} M \times N$ and $f \in C^{\infty}M \times N$, how do you intepret $v\circ f$ interms of $T_pM \oplus T_q N$?

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Suppose you choose a basis for $T_p M$, given by the derivatives $\partial_i$, and a basis for $T_q N$, given by the derivatives $\partial_a$. Here, $i$ runs over the $M$ indexes, while $a$ runs over the $N$ indexes.

A vector in $T_p M\oplus T_q N$ can be decomposed in components as $$ v=v^i\partial_i+v^a\partial_a, $$ so it acts on the functions $f$ of $(p,q)$ according to $$ v(f(p,q))=v^i(\partial_if(p,q))+v^a(\partial_a f(p,q)), $$ where $(p,q)$ is expressed in coordinates.

As an example, take $\mathbb R^2\times \mathbb R$ with standard metric, take the standard cartesian coordinates and adopt the basis $\{\partial_x,\partial_y\}$ over $T_p \mathbb R^2$ and $\{\partial_z\}$ over $T_q \mathbb R$ (x and y are coordinates on $\mathbb R^2$, z is a coordinate on $\mathbb R$). The vector of coordinates over $T_p \mathbb R^2\oplus T_q\mathbb R$ given by the coordinates $(1,0,1)$ is $$ v=\partial_x+\partial_z. $$ Take the test function $f(x,y,z)=\sin(xyz^2)$: $v$ acts on $f$ giving the function $$ (v\circ f)(x,y,z)=\partial_x f(x,y,z)+\partial_z f(x,y,z)=yz^2\cos(xyz^2)+2xyz\cos(xyz^2). $$

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    $\partial_i \in T_pM$, how does it act on $f \in C^{\infty}(M \times N)$? – 2017-02-02
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    As I wrote, it acts as a partial derivative with respecto to the appropriate coordinate on $(M\times N)$. See the example above. – 2017-02-02