The other answer is very helpful, but I believe it has some subtle problems. Let me expand on a few of the details because I think they can be confusing.
The tangent bundle $T E$ of a vector bundle $p : E \to B$ is canonically itself a vector bundle in two different ways: over $E$ via the usual $\pi_E : T E \to E$, and via $D p : TE \to TB$.
The vertical subbundle $V E \subset T E$ is the kernel of $D p$, and so is itself a vector bundle over $E$ (but not $T B$). Also, let's call the kernel of $\pi_E : T E \to E$ (where we give $T E$ the $TB$-bundle structure) the zero subbundle $Z E$ of $T E$. Then $Z E$ is a vector bundle over $TB$ (but not $E$). Of course, $ZE$ is the same as the restriction of $TE$ with the usual $E$-bundle structure to the image of the zero section $\sigma(B) \subset E$.
In this language, a linear connection on $E$ is the same as a choice of $E$-bundle complement $HE$ of the subbundle $VE$ in $TE$ such that $HE$ is at the same time a $TB$-bundle complement of $ZE$. This is a bit confusing: $HE$ is simultaneously a vector subbundle of $TE$ with respect to both possible bundle structures.
The horizontal subbundle $H(E \otimes F) \subset T(E \otimes F)$
Thus in the presence of a connection, we have direct sum decompositions $TE \cong VE \oplus_E HE$ and $TE \cong ZE \oplus_{TB} HE$. Now consider the map of both $(E\otimes F)$-bundles and $TB$-bundles
$$
\tau : TE \otimes_{TB} TF \to T(E \otimes_B F).
$$
The bundle $TE \otimes_{TB} TF$ has twice the fiber dimension of $T(E \otimes_B F)$, so $\tau$ must have a kernel. This is annoying to pin down, since e.g. the image of the tensor product of vertical vectors
$$
\tau : [e + t e'] \otimes [f + t f'] \mapsto [e \otimes f + t(e \otimes f' + e' \otimes f)]
$$
has a funky multiplication law. It turns out that the tensor product plays much more nicely with the decomposition $TE \cong ZE \oplus_{TB} HE$ : if $\alpha \in ZE$ and $\beta \in ZF$ then $\tau(\alpha \otimes \beta) = 0$ (intuitively because $\alpha \otimes \beta$ must be zero "to second order"). Also, we pretty easily see that $\tau(HE \otimes ZF) = \tau(ZE \otimes HF) = Z(E \otimes F)$ and in either case the corresponding restriction of $\tau$ is an isomorphism onto its image.
By counting dimensions, this completely pins down the kernel of $\tau$: it is the direct sum $K_0 \oplus K_1$ with $K_0 = ZE \otimes ZF$ and $K_1$ spanned by all $\gamma = \alpha_H \otimes \beta_Z + \alpha_Z \otimes \beta_H$ with $\alpha_H \in HE$, $\alpha_Z \in ZE$, $\beta_H \in HF$, $\beta_Z \in ZF$, and $\tau(\gamma) = 0$.
In particular, for dimension reasons this means that $\tau$ takes $HE \otimes HF$ isomorphically onto its image. We then define the horizontal subbundle $H(E \otimes F) := \tau(HE \otimes HF)$ of $E \otimes F$ this way. It's easy to see that $H(E \otimes F)$ is a complement to $Z(E \otimes F)$ in the right sense. Thus we have obtained the canonical connection on the tensor product bundle $E \otimes F$.
The horizontal projection $h_{E \otimes F} : T (E \otimes F) \to T (E \otimes F)$
Let $h_E : T E \to T E$ and $h_F : T F \to T F$ be the horizontal projections for conenctions on $E$ and $F$ respectively. Because $T (E \otimes F)$ is awkward to work with directly, it is easiest define the horizontal projection $h_{E \otimes F} : T (E \otimes F) \to T (E \otimes F)$ by first building the map
$$
TE \otimes_{TB} TF \xrightarrow{h_E \otimes h_F} TE \otimes_{TB} TF \xrightarrow{\ \ \ \ \tau\ \ \ \ } T(E \otimes F).
$$
This map descends to a map $h_{E \otimes F} : T(E \otimes F) \to T(E \otimes F)$ exactly when $\tau \eta = 0$ implies $\tau (h_E \otimes h_F) \eta = 0$ for any $\eta \in TE \otimes_{TB} TF$. But this is true by our explicit calculations above: the kernel of $\tau$ is $K_0 \oplus K_1$, clearly $(h_E \otimes h_F)K_0 = 0$, and for any $\gamma = \alpha_H \otimes \beta_Z + \alpha_Z \otimes \beta_H \in K_1$ we have
$$
(h_E \otimes h_F)\gamma = \alpha_H \otimes 0 + 0 \otimes \beta_H = 0.
$$
Thus our map descends and we obtain the horizontal projection map $h_{E \otimes F}$ satisfying for any $\alpha$ and $\beta$
$$
h_{E \otimes F}(\tau(\alpha \otimes \beta)) = \tau(h_E \alpha \otimes h_F \beta).
$$
The connector $K_{E \otimes F} : T (E \otimes F) \to E \otimes F$
Given the connectors $K_E : T E \to E$ and $K_F : T F \to F$, what is the connector $K_{E \otimes F}$? Tracing through the definition of the connector in terms of $h$ we obtain
$$
K_{E \otimes F} \circ \tau =
K_E \otimes \pi_F + \pi_E \otimes K_F,
$$
which is subtly different to Peter's formula (as I explain in comments).
If you want to compute this formula yourself note the following (and replace $E$ with $E \otimes F$ below): usually the connector $K_E$ of a connection is defined as the composite
$$
T E \xrightarrow{\ \ \operatorname{vpr}\ \ } V E \xrightarrow{\operatorname{vlift}^{-1}} E \times_B E \xrightarrow{\ \ \operatorname{pr}_2\ \ } E
$$
where $\operatorname{vpr} : T E \to V E$ is the vertical projection and $\operatorname{vlift} : E \times_B E \to V E$ is the vertical lift isomorphism. Here $\operatorname{vpr} = 1 - h_E$ with the subtraction taken with respect to the $E$-bundle structure. It is much more convenient in our language to define the connector as the composite
$$
T E \xrightarrow{\ \ \operatorname{zpr}\ \ } Z E \xrightarrow{\operatorname{zlift}^{-1}} E \times_B E \xrightarrow{\ \ \operatorname{pr}_2\ \ } E
$$
where $\operatorname{zpr} : T E \to Z E$ is the projection onto $ZE$ and $\operatorname{zlift} : TB \times_B E \to ZE$ is the "zero lift" isomorphism. Here $\operatorname{zpr} = 1 - h_E$ with the subtraction taken with respect to the $TB$-bundle structure. As an exercise you can prove that these two definitions of $K_E$ are equivalent.
If you are not familiar with the map I have called "$\operatorname{zlift}$", its inverse $ZE \to TB \times_B E$ is defined by (if $p : E \to B$ is a vector bundle and $\xi \in ZE$ is represented by the derivative of the curve $\gamma(t) : \mathbb{R} \to E$ at $t = 0$)
$$
\xi \mapsto \left(Dp(X),\ \lim_{t \to 0} \frac{\gamma(t)}{t}\right).
$$
Note that the limit always exists exactly because $\xi \in ZE$ and so $\gamma(0) = \pi_E(\xi) = 0$.
General situation
In my math.se answer here I explain that in general there is a canonical isomorphism
$$
\mu := \tau \circ (\operatorname{hlift} \otimes \operatorname{id}_{TF}) : (TB \times E) \otimes_{TB} TF \to T(E \otimes F)
$$
given just a connection on $E$ only. Here $\operatorname{hlift} : TB \times_B E \to HE$ is the horizontal lift isomorphism for the connection, and of course my notation $(TB \times E) \otimes_{TB} TF$ means the same as Peter's $E \otimes TF$. Under the identification $\mu$ I compute that if one also chooses a connection on $F$ then $K_{E \otimes F} : (TB \times E) \otimes TF \to E \otimes F$ is just $\operatorname{pr}_2 \otimes K_F$, with the connection on $E$ hidden inside the identification $\mu$ itself.
