If we define the map $f\times g : P\to M\times N$ by $f\times g : x \mapsto (f(x),g(x))$ and if we define the maps $i_p : N \to M\times N$ by $i_p(n)=(p,n)$ and $i_q : M \to M\times N$ by $i_q(m)=(m,q)$ then is it true that for the differential we have $d(f\times g)_x=(di_{g(x)})_{f(x)}\circ df_x + (di_{f(x)})_{g(x)}dg_x$?
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Following the link that Eumenes suggested, since $d(f\times g)_x(v) \in T_{(f(x),g(x))}M\times N$ we get \begin{align} \sigma(d(f\times g)_x(v))&=(d\pi_1(d(f\times g)_x(v)),d\pi_2(d(f\times g)_x(v)))\\ &=(d(\pi_1\circ f\times g)_x(v), d(\pi_2\circ f\times g)_x(v))\\ &=(df_x(v),dg_x(v)) \end{align}
So then
$\tau(\sigma(d(f\times g)_x(v))=(di_{g(x)})_{f(x)}(df_x(v))+(di_{f(x)})_{g(x)}(dg_x)(v)$
But it is kind of weird to me that we don't get the same expression at the end, since $\sigma$ and $\tau$ are inverse maps so $\tau \circ \sigma=Id$
Let's check this on the level of coordinates (although you can see this without resorting to coordinates). Let coordinates on $P$ be $(a_1,...,a_j)$, on $M$ be $(x_1,...,x_k)$ and on $N$ be $(y_1,...,y_l)$. This yields coordinates $(x_1,...,x_k,y_1,...,y_l)$. Then $D(f\times g)_a$ looks like the matrix for $Df_a$ on top of $Dg_a$. Now note that $Di_q$ is a $(k+l)\times k$ matrix whose upper $k \times k$ block is the identity, and zeros elsewhere. Similarly $Di_p$ is a $(k+l) \times l$ matrix whose lower $l \times l$ matrix is the identity, and zeros elsewhere. As a result, $(Di_q)_{f(a)}Df_a + (Di_p)_{g(a)}Dg_a$ exactly "slots" $Df_a$ on top and $Dg_a$ on the bottom.
One last note. I don't think this makes sense unless you rephrase the last line as $$d(f \times g)_x = (di_{g(x)})_{f(x)}\circ df_x + (di_{f(x)})_{g(x)}\circ dg_x$$ because otherwise you aren't adding tangent vectors at the same point $(f(x),g(x))$. This sort of reflects the issue that there's no canonical inclusion of $M$ or $N$ into $M \times N$. However, you can instead "reverse the arrows" and consider the situation where, you look at the projections $\pi_M:M \times N \to M$ and $\pi_N: M\times N \to N$. There is no ambiguity here. In fact, $T_{(m,n)}(M \times N)\cong T_m(M) \times T_n(N)$, and the universal property of products say that if you have maps $df_p:T_p(P) \to T_m(M)$ and $dg_q:T_p(P) \to T_n(N)$ then you have a unique map to $d(f\times g)_{(p,q)}:P \to T_m(M) \times T_n(N)$ such that it commutes with these projections. So (at least on the level of tangent spaces, and it is true on the level of maps) a smooth map to the product is determined by a smooth map to each factor.