In the second spectral sequence you write $R^q f_\ast I^\bullet$. This is correct but $R^qf_\ast$ must be considered as the $q$th hyper-derived functor applied to the complex $I^\bullet$, it is not the ordinary derived functor $R^qf_\ast$ applied termwise. See my answer to Construction of the spectral sequence of Katz/Oda for more details.

Once you observe this, then you find that the hypercohomology spectral sequence is just the special case of the Leray spectral sequence when $f = \mathrm{id}$. Indeed the hyper-derived functor $R^q \mathrm{id}_\ast$ is just the functor $\mathcal H^q$.

Addendum: One situation where what you want is actually true (and which is more general than $f=\mathrm{id}$) is if all cohomology sheaves $\mathcal H^q(I^\bullet)$ are $f_\ast$-acyclic, i.e. if they vanish under $R^qf_\ast$ for $q> 0$. This implies that $$ R^qf_\ast I^\bullet \cong f_\ast \mathcal H^q I^\bullet$$ (where again on the left hand side I mean the hyper-derived functor), and also that $$ H^p(X,\mathcal H^q I^\bullet) \cong H^p(X',f_\ast\mathcal H^q I^\bullet).$$It's not hard to show that there is even an isomorphism between the two spectral sequences.

But this is a very restrictive condition. What you're asking for is not very natural -- in general there is not even a map between the two spectral sequences.

In the second spectral sequence you write $R^q f_\ast I^\bullet$. This is correct but $R^qf_\ast$ must be considered as the $q$th hyper-derived functor applied to the complex $I^\bullet$, it is not the ordinary derived functor $R^qf_\ast$ applied termwise. See my answer to Construction of the spectral sequence of Katz/Oda for more details.

Once you observe this, then you find that the hypercohomology spectral sequence is just the special case of the Leray spectral sequence when $f = \mathrm{id}$. Indeed the hyper-derived functor $R^q \mathrm{id}_\ast$ is just the functor $\mathcal H^q$.

Addendum: One situation where what you want is actually true (and which is more general than $f=\mathrm{id}$) is if all cohomology sheaves $\mathcal H^q(I^\bullet)$ are $f_\ast$-acyclic, i.e. if they vanish under $R^qf_\ast$ for $q> 0$. This implies that $$ R^qf_\ast I^\bullet \cong f_\ast \mathcal H^q I^\bullet$$ (where again on the left hand side I mean the hyper-derived functor), and also that $$ H^p(X,\mathcal H^q I^\bullet) \cong H^p(X',f_\ast\mathcal H^q I^\bullet).$$It's not hard to show that there is even an isomorphism between the two spectral sequences.

But this is a very restrictive condition. What you're asking for is not very natural -- in general there is not even a map between the two spectral sequences.

In the second spectral sequence you write $R^q f_\ast I^\bullet$. This is correct but $R^qf_\ast$ must be considered as the $q$th hyper-derived functor applied to the complex $I^\bullet$, it is not the ordinary derived functor $R^qf_\ast$ applied termwise. See my answer to Construction of the spectral sequence of Katz/Oda for more details.

Once you observe this, then you find that the hypercohomology spectral sequence is just the special case of the Leray spectral sequence when $f = \mathrm{id}$. Indeed the hyper-derived functor $R^q \mathrm{id}_\ast$ is just the functor $\mathcal H^q$.

In the second spectral sequence you write $R^q f_\ast I^\bullet$. This is correct but $R^qf_\ast$ must be considered as the $q$th hyper-derived functor applied to the complex $I^\bullet$, it is not the ordinary derived functor $R^qf_\ast$ applied termwise. See my answer to Construction of the spectral sequence of Katz/Oda for more details.

Once you observe this, then you find that the hypercohomology spectral sequence is just the special case of the Leray spectral sequence when $f = \mathrm{id}$. Indeed the hyper-derived functor $R^q \mathrm{id}_\ast$ is just the functor $\mathcal H^q$.

Addendum: One situation where what you want is actually true (and which is more general than $f=\mathrm{id}$) is if all cohomology sheaves $\mathcal H^q(I^\bullet)$ are $f_\ast$-acyclic, i.e. if they vanish under $R^qf_\ast$ for $q> 0$. This implies that $$ R^qf_\ast I^\bullet \cong f_\ast \mathcal H^q I^\bullet$$ (where again on the left hand side I mean the hyper-derived functor), and also that $$ H^p(X,\mathcal H^q I^\bullet) \cong H^p(X',f_\ast\mathcal H^q I^\bullet).$$It's not hard to show that there is even an isomorphism between the two spectral sequences.

But this is a very restrictive condition. What you're asking for is not very natural -- in general there is not even a map between the two spectral sequences.