$d(f \times g)_{x,m} = df_x \times dg_m$? 
(a) $d(f \times g)_{x,m} = df_x \times dg_m$?

Also, 

(b) does $d(f \times g)_{x,m}$ carry $\tilde{x} \in T_x X, \tilde{m} \in T_x M$ to the tangent space of $f$ cross the tangent space of $g$?

Furthermore,

(c) does $df_x \times dg_m$ carry $\tilde{x} \in T_x X, \tilde{m} \in T_x M$ to the tangent space of $f$ cross the tangent space of $g$?

Thank you~

I don't know what formalism you are adopting, but starting from this one it can be done intuitively. I will state the construction without proof of compatibility and so.
An element of $T_x X$ is a differential from locally defined functions $C^{1}_x(X)$ to $\mathbb{R}$ induced by a locally defined curve $\gamma_x(t):(-\epsilon,\epsilon)\rightarrow X,\gamma_x(0)=x$ by $\gamma_x (f)=\frac{df(\gamma(t))}{dt}|_{t=0}$.
$T_x X\times T_m M$ is identified canonically with $T_{(x,m)}(X\times M)$ in the following way. Given $\gamma_x,\gamma_m$, then $\gamma_{(x,m)}(t)=(\gamma_x(t),\gamma_m(t))$; given $\gamma_{(x,m)}$ then $\gamma_x(t)=p_X\gamma_{x,m}(t)$ where $p_X$ is the projection, and likewise for $m$.
For a map $F:X\rightarrow Y$, the differential is defined as $dF_x(\gamma_x)(g)=\gamma_x(gF)$.
Now what you want is a commutative diagram
\begin{array}{ccc}
T_{(x,m)}(X\times M) & \cong & T_x X\times T_m M \\
\downarrow &   & \downarrow \\
T_{(y,n)}(Y\times N) & \cong & T_y Y\times T_n N \end{array}
Let's now check it. $\gamma_{(x,m)}$ is mapped to $\gamma_{(x,m)}(g(F,G))$, by definition of the tangent vector, we see it is induced by the curve $(F,G)\gamma_{(x,m)}=(Fp_X \gamma_{(x,m)},Gp_M \gamma_{(x,m)})$, which then projects to $Fp_X \gamma_{(x,m)}$ and $Gp_M \gamma_{(x,m)}$. If we factored earlier, it would first be taken to $p_X \gamma_{(x,m)}$ and $p_M \gamma_{(x,m)}$, then by $F,G$ respectively to $Fp_X \gamma_{(x,m)}$ and $Gp_M \gamma_{(x,m)}$, which is the same.
