For many years I have sneered at formal categorical arguments, but recently I have started appreciating their elegance, even though I'm still quite inexperienced.
I believe that my question has a very elementary answer, but I have drawn many diagrams so far without arriving at any conclusive answer.
Let $E\to Y$ and $F\to Y$ be two vector bundles and let $f\colon X\to Y$ be a continuous map. I would like to show, using only universal properties, that $f^*(E\oplus F)$ and $f^*(E\otimes F)$ are naturally isomorphic to $(f^*E) \oplus (f^*F)$ and $(f^*E)\otimes (f^*F)$ respectively.
I start out with the two diagrams $\require{AMScd}$ \begin{CD} E\oplus F @>>> E\\ @VVV @VVV\\ F @>>> Y \end{CD} and \begin{CD} f^*(E\oplus F) @>>> E\oplus F \\ @VVV @VVV\\ X @>f>> Y \end{CD} but I'm unable to compose them in a suitable way with \begin{CD} f^*E @>>> E \\ @VVV @VVV\\ X @>f>> Y \end{CD} and \begin{CD} f^*F @>>> F \\ @VVV @VVV\\ X @>f>> Y \end{CD} to obtain the desired natural isomorphism. And for the tensor product, I'm even further from an answer.
I think one way to do this is to rely on the fact that if $g:A\rightarrow C$ and $h:B\rightarrow D$, with bundles $E\rightarrow C$ and $F\rightarrow D$, then $$ (g\times h)^\ast(E\times F) \cong g^\ast(E)\times h^\ast(F)$$
as bundles on $A\times B$. I don't know a categorical argument for this (but I'm sure one exists), but the proof is very easy using the universal property of the pullback.
Now, the Whitney sum can be done as follows. If $g:A\rightarrow C$ and $E,F$ are bundles over $C$, then $$ \Delta\circ g\equiv (g\times g)\circ\Delta $$
as maps from $A$ to $C\times C$, where $\Delta:X\rightarrow X\times X$ is the diagonal map. Now all that's left to do is use the first displayed equation, and the fact the Whitney sum is the pullback of the diagonal map. I believe the tensor product can be done similarly.