Yes, that's true. Let $f_i : V_i \to W_i$ be two linear maps. Since $\mathrm{im}(f_1) \otimes \mathrm{im}(f_2)$ embeds into $W_1 \otimes W_2$, we may assume that $f_1,f_2$ are surjective. But then they are split, so that we can assume that $V_i = W_i \oplus U_i$ and that $f_i$ equals the projection $V_i \to W_i$, with kernel $U_i$. Then $V_1 \otimes V_2 = W_1 \otimes W_2 \oplus W_1 \otimes U_2 \oplus U_1 \otimes W_2 \oplus U_1 \otimes U_2$ and $f_1 \otimes f_2$ equals the projection of $V_1 \otimes V_2$ onto $W_1 \otimes W_2$. Hence the kernel is $W_1 \otimes U_2 \oplus U_1 \otimes W_2 \oplus U_1 \otimes U_2 = U_1 \otimes V_2 + V_1 \otimes U_2$.
This shows even more: The kernel is the pushout $(\ker(f_1) \otimes V_2) \cup_{\ker(f_1) \otimes \ker(f_2)} (V_1 \otimes \ker(f_2))$.
By the way, this argument is purely formal and works in every semisimple abelian $\otimes$-category. What happens when we drop semisimplicity, for example when we consider modules over some commutative ring $R$? Then we only need some flatness assumptions:
Let $f_1 : V_1 \to W_1$ and $f_2 : V_2 \to W_2$ be two morphisms in an abelian $\otimes$-category (for example $R$-linear maps between $R$-modules). If $f_1,f_2$ are epimorphisms, then we have exact sequences $\ker(f_1) \to V_1 \to W_1 \to 0$ and $\ker(f_2) \to V_2 \to W_2 \to 0$. Applying the right exactness of the tensor product twice(!), we get that then also
$\ker(f_1) \otimes V_2 \oplus V_1 \otimes \ker(f_2) \to V_1 \otimes V_2 \to W_1 \otimes W_2 \to 0$
is exact. If $f_1,f_2$ are not epi, we can still apply the above to their images and get the exactness of
$\ker(f_1) \otimes V_2 \oplus V_1 \otimes \ker(f_2) \to V_1 \otimes V_2 \to \mathrm{im}(f_1) \otimes \mathrm{im}(f_2) \to 0.$
Now assume that $\mathrm{im}(f_1)$ and $W_2$ are flat. Then $\mathrm{im}(f_1) \otimes \mathrm{im}(f_2)$ embeds into $\mathrm{im}(f_1) \otimes W_2$ which embeds into $W_1 \otimes W_2$. Hence, we have still that the sequence
$\ker(f_1) \otimes V_2 \oplus V_1 \otimes \ker(f_2) \to V_1 \otimes V_2 \to W_1 \otimes W_2$
is exact. In other words, we have a sum decomposition
$$\ker(f_1 \otimes f_2) = \alpha(\ker(f_1) \otimes V_2) + \beta(V_1 \otimes \ker(f_2)),$$ where $\alpha : \ker(f_1) \otimes V_2 \to V_1 \otimes V_2$ and $\beta : V_1 \otimes \ker(f_2) \to V_1 \otimes V_2$ are the canonical morphisms. In general, these are not monic! However, this is the case, by definition, when $V_1$ and $V_2$ are flat. So in this case we can safely treat $\alpha$ and $\beta$ as inclusions and write $$\ker(f_1 \otimes f_2) = V_1 \otimes \ker(f_2) + \ker(f_1) \otimes V_2.$$
Take
$$\omega := \phi^1\wedge \phi^2 = \phi^1\otimes \phi^2 - \phi^2 \otimes \phi^1,$$
which is an element of $\Lambda^2(T^*W)$. Its components are found by feeding it $f_i$'s. For the pullback, which is an element of $\Lambda^2(T^*V)$, the components are found (or rather defined) by feeding it $e_i$'s as prescribed by the formula:
$$\big(L^*\omega\big)_{ij} = L^*\omega(e_i,e_j) := \omega(L(e_i),L(e_j))$$
This differs in appearance from your formula, but the idea is the same (push forward args and apply omega).
Explicitly, we have:
$$L(e_1) = f_2,\text{ }\text{ }\text{ }L(e_2) = f_1,\text{ }\text{ }\text{ }L(e_3)=f_1+f_2,$$
Hence for example:
$$\color{blue}{(L^*\omega)_{11}} = (\phi^1\otimes\phi^2 - \phi^2\otimes\phi^1)(L(e_1),L(e_1))$$
$$= \phi^1(f_2)\cdot\phi^2(f_2) - \phi^2(f_2)\cdot\phi^1(f_2) = \delta^1_2\cdot \delta^2_2 - \delta^2_2\cdot\delta^1_2 = 0-0 = \color{blue}{0}$$
($\delta^i_j$ is the Kronecker delta). The other eight components are found similarly.
I would like to add that this is a skew-symmetric tensor, so you need only find $(L^*\omega)_{12}$, $(L^*\omega)_{13}$, and $(L^*\omega)_{23}$ since diagonal elements are all zero and the rest are negatives of the above.
Best Answer
If $V$ is a $\Bbb C$-vector space, then we can define a conjugate vector space $c(V)$ as follows.
The underlying abelian group of $c(V)$ is the same as the abelian group $V$. The scalar product $\bullet$ in $c(V)$ is defined as $\alpha \bullet v = \overline\alpha \cdot v$, where $\cdot$ denotes the scalar product on $V$.
Thus by definition, the map $f_1$ is nothing but a $\Bbb C$-linear map from $X_1$ to $c(Y_1)$. Same for $f_2$.
Now it only remains to prove a
Lemma: $c(Y_1) \otimes c(Y_2)$ is canonically isomorphic to $c(Y_1 \otimes Y_2)$ as $\Bbb C$-vector spaces.
The proof of the lemma is a simple exercise, and the original question immediately follows from the lemma.