Let $X$ be ascheme and $f :X \to \mathbb{P}^n$ a morphism such that it gives rise for the exact sequence
$$ 0 \longrightarrow \mathcal{I}_X \longrightarrow \mathcal{O}_{\mathbb{P}^n} \longrightarrow f_*\mathcal{O}_X \longrightarrow 0 ~, $$
for example if $f =\imath:X \hookrightarrow \mathbb{P}^n$ is an embedding.
where $\mathcal{I}$ is the ideal sheaf of $X$ (the sheaf of sections of $\mathcal{O}_{\mathbb{P}^n}$ which vanish on $X$).
Now we can tensoring this by $\mathcal{O}_{\mathbb{P}^n}(d)$ to get $$ 0 \longrightarrow \mathcal{I}_X(d) \longrightarrow \mathcal{O}_{\mathbb{P}^n}(d) \longrightarrow f_*\mathcal{O}_X \otimes \mathcal{O}_{\mathbb{P}^n}(d) \longrightarrow 0 $$
But I read often that tensoring induces to sequence $$ 0 \longrightarrow \mathcal{I}_X(d) \longrightarrow \mathcal{O}_{\mathbb{P}^n}(d) \longrightarrow f_*\mathcal{O}_X(d) \longrightarrow 0 $$
This would imply that $f_*\mathcal{O}_X \otimes \mathcal{O}_{\mathbb{P}^n}(d) = f_*\mathcal{O}_X(d)$. Can anybody explain why this equality holds?
This follows from applying the projection formula (Hartshorne II.5, Exercise 5.1d) to the $\mathcal O_X$-module $\mathcal O_X$, and the locally-free finite-rank $\mathcal O_{\mathbb P^n}$-module $\mathcal O_{\mathbb P^n}(d)$: $$ i_\star \left( \mathcal O_X \otimes_{\mathcal O_X}i^\star\mathcal O_{\mathbb P^n}(d) \right) \cong i_\star \mathcal O_X \otimes_{\mathcal O_{\mathbb P^n}}\mathcal O_{\mathbb P^n}(d)$$
But since $i : X \hookrightarrow \mathbb P^n$ is a projective embedding, $i^\star\mathcal O_{\mathbb P^n}(d)$ is the same as $\mathcal O_X(d)$, by the definition of $\mathcal O_X(d)$. [Note that it is important that the $i$ is a projective embedding, and the invertible sheaf $\mathcal O_X(d)$ is defined with respect to this particular projective embedding $i$.]
Finally, $\mathcal O_X \otimes_{\mathcal O_X}\mathcal O_{X}(d) $ is obviously the same thing as $\mathcal O_{X}(d)$. So putting everything together, we get your desired identity, $$ i_\star \mathcal O_X(d)\cong i_\star \mathcal O_X \otimes_{\mathcal O_{\mathbb P^n}} \mathcal O_{\mathbb P^n}(d). $$