Positive-definiteness of the curvature operator ($R>0$) is as much stronger condition than positive sectional curvature ($\sec>0$). In fact, as Igor mentions, it follows from the work of Boehm and Wilking that only spherical space forms admit metrics with $R>0$, while many (but not so many!) other manifolds admit metrics with $\sec>0$. An almost complete list of such manifolds can be found in Section 4 of this survey of Ziller.

Of course, what happens is that although $R\colon\wedge^2 T_pM\to\wedge^2 T_pM$ is symmetric (hence diagonalizable), its eigenspaces need not intersect the Grassmannian $Gr_2(T_pM)=\{\sigma\in\wedge^2 T_pM:\sigma\wedge\sigma=0,\|\sigma\|=1\}$. Hence $R$ could have some zero (or even negative) eigenvalues at the same time as the restriction $\sec(\sigma)=\langle R(\sigma),\sigma\rangle$ of its quadratic form to $Gr_2(T_pM)$ is positive. For example, in $\mathbb C P^n$ this is what happens: $R$ is positive-semidefinite (and has nontrivial kernel), but this kernel does not intersect the Grassmannian.

Algebraically, there is an intermediate curvature condition between $\sec>0$ and $R>0$ called strongly positive curvature, that might be of interest to you. Namely, a curvature operator $R\colon\wedge^2 V\to\wedge^2 V$ has strongly positive curvature if there exists a $4$-form $\omega\in\wedge^4V$ such that $R+\omega$ is positive-definite. Here, $\omega\in\wedge^4 V$ is identified with a symmetric endomorphism $\omega\colon\wedge^2V\to\wedge^2V$ via $$\langle\omega(\alpha),\beta\rangle=\langle\omega,\alpha\wedge\beta\rangle.$$ Clearly, $R>0$ implies strongly positive curvature (take $\omega=0$). Since $\sec(\sigma)=\langle R(\sigma),\sigma\rangle$ and $\langle\omega(\sigma),\sigma\rangle=\langle\omega,\sigma\wedge\sigma\rangle=0$ if $\sigma$ is decomposable, strongly positive curvature implies $\sec>0$. Together with R. Mendes, over the last years, I have pursued a systematic study of this curvature condition; see the following for details:  https://arxiv.org/abs/1403.2117, https://arxiv.org/abs/1412.0039,  https://arxiv.org/abs/1511.07899. For example, $(\mathbb C P^n,g_{FS})$ has strongly positive curvature, since $R+\varepsilon (\omega_{FS}\wedge\omega_{FS})>0$ for small $\varepsilon>0$, where $\omega_{FS}$ is the Fubini-Study $2$-form.

The upshot is that almost all known examples of manifolds with $\sec>0$ actually satisfy this much stronger condition. For manifolds with $\sec\geq0$, actually all known examples satisfy the analogous "strongly nonnegative curvature" condition (which requires $R+\omega$ to be positive-semidefinite).

Positive-definiteness of the curvature operator ($R>0$) is as much stronger condition than positive sectional curvature ($\sec>0$). In fact, as Igor mentions, it follows from the work of Boehm and Wilking that only spherical space forms admit metrics with $R>0$, while many (but not so many!) other manifolds admit metrics with $\sec>0$. An almost complete list of such manifolds can be found in Section 4 of this survey of Ziller.

Of course, what happens is that although $R\colon\wedge^2 T_pM\to\wedge^2 T_pM$ is symmetric (hence diagonalizable), its eigenspaces need not intersect the Grassmannian $Gr_2(T_pM)=\{\sigma\in\wedge^2 T_pM:\sigma\wedge\sigma=0,\|\sigma\|=1\}$. Hence $R$ could have some zero (or even negative) eigenvalues at the same time as the restriction $\sec(\sigma)=\langle R(\sigma),\sigma\rangle$ of its quadratic form to $Gr_2(T_pM)$ is positive. For example, in $\mathbb C P^n$ this is what happens: $R$ is positive-semidefinite (and has nontrivial kernel), but this kernel does not intersect the Grassmannian.

Algebraically, there is an intermediate curvature condition between $\sec>0$ and $R>0$ called strongly positive curvature, that might be of interest to you. Namely, a curvature operator $R\colon\wedge^2 V\to\wedge^2 V$ has strongly positive curvature if there exists a $4$-form $\omega\in\wedge^4V$ such that $R+\omega$ is positive-definite. Here, $\omega\in\wedge^4 V$ is identified with a symmetric endomorphism $\omega\colon\wedge^2V\to\wedge^2V$ via $$\langle\omega(\alpha),\beta\rangle=\langle\omega,\alpha\wedge\beta\rangle.$$ Clearly, $R>0$ implies strongly positive curvature (take $\omega=0$). Since $\sec(\sigma)=\langle R(\sigma),\sigma\rangle$ and $\langle\omega(\sigma),\sigma\rangle=\langle\omega,\sigma\wedge\sigma\rangle=0$ if $\sigma$ is decomposable, strongly positive curvature implies $\sec>0$. Together with R. Mendes, over the last years, I have pursued a systematic study of this curvature condition; see the following for details:

Bettiol, Renato G.; Mendes, Ricardo A. E., Strongly positive curvature, Ann. Global Anal. Geom. 53, No. 3, 287-309 (2018). ZBL1395.53058. (arXiv)

Bettiol, Renato G.; Mendes, Ricardo A. E., Strongly nonnegative curvature, Math. Ann. 368, No. 3-4, 971-986 (2017). ZBL1377.53020. (arXiv)

Bettiol, Renato G.; Mendes, Ricardo A. E., Flag manifolds with strongly positive curvature, Math. Z. 280, No. 3-4, 1031-1046 (2015). ZBL1360.53056. (arXiv)

For example, $(\mathbb C P^n,g_{FS})$ has strongly positive curvature, since $R+\varepsilon (\omega_{FS}\wedge\omega_{FS})>0$ for small $\varepsilon>0$, where $\omega_{FS}$ is the Fubini-Study $2$-form.

The upshot is that almost all known examples of manifolds with $\sec>0$ actually satisfy this much stronger condition. For manifolds with $\sec\geq0$, actually all known examples satisfy the analogous "strongly nonnegative curvature" condition (which requires $R+\omega$ to be positive-semidefinite).

Positive-definiteness of the curvature operator ($R>0$) is as much stronger condition than positive sectional curvature ($\sec>0$). In fact, as Igor mentions, it follows from the work of Boehm and Wilking that only spherical space forms admit metrics with $R>0$, while many (but not so many!) other manifolds admit metrics with $\sec>0$. An almost complete list of such manifolds can be found in Section 4 of this survey of Ziller.

Of course, what happens is that although $R\colon\wedge^2 T_pM\to\wedge^2 T_pM$ is symmetric (hence diagonalizable), its eigenspaces need not intersect the Grassmannian $Gr_2(T_pM)=\{\sigma\in\wedge^2 T_pM:\sigma\wedge\sigma=0,\|\sigma\|=1\}$. Hence $R$ could have some zero (or even negative) eigenvalues at the same time as the restriction $\sec(\sigma)=\langle R(\sigma),\sigma\rangle$ of its quadratic form to $Gr_2(T_pM)$ is positive. For example, in $\mathbb C P^n$ this is what happens: $R$ is positive-semidefinite (and has nontrivial kernel), but this kernel does not intersect the Grassmannian.

Algebraically, there is an intermediate curvature condition between $\sec>0$ and $R>0$ called strongly positive curvature, that might be of interest to you. Namely, a curvature operator $R\colon\wedge^2 V\to\wedge^2 V$ has strongly positive curvature if there exists a $4$-form $\omega\in\wedge^4V$ such that $R+\omega$ is positive-definite. Here, $\omega\in\wedge^4 V$ is identified with a symmetric endomorphism $\omega\colon\wedge^2V\to\wedge^2V$ via $$\langle\omega(\alpha),\beta\rangle=\langle\omega,\alpha\wedge\beta\rangle.$$ Clearly, $R>0$ implies strongly positive curvature (take $\omega=0$). Since $\sec(\sigma)=\langle R(\sigma),\sigma\rangle$ and $\langle\omega(\sigma),\sigma\rangle=\langle\omega,\sigma\wedge\sigma\rangle=0$ if $\sigma$ is decomposable, strongly positive curvature implies $\sec>0$. Together with R. Mendes, over the last years, I have pursued a systematic study of this curvature condition; see the following for details: https://arxiv.org/abs/1403.2117, https://arxiv.org/abs/1412.0039, https://arxiv.org/abs/1511.07899.

The upshot is that almost all known examples of manifolds with $\sec>0$ actually satisfy this much stronger condition. For manifolds with $\sec\geq0$, actually all known examples satisfy the analogous "strongly nonnegative curvature" condition (which requires $R+\omega$ to be positive-semidefinite).

Positive-definiteness of the curvature operator ($R>0$) is as much stronger condition than positive sectional curvature ($\sec>0$). In fact, as Igor mentions, it follows from the work of Boehm and Wilking that only spherical space forms admit metrics with $R>0$, while many (but not so many!) other manifolds admit metrics with $\sec>0$. An almost complete list of such manifolds can be found in Section 4 of this survey of Ziller.

Of course, what happens is that although $R\colon\wedge^2 T_pM\to\wedge^2 T_pM$ is symmetric (hence diagonalizable), its eigenspaces need not intersect the Grassmannian $Gr_2(T_pM)=\{\sigma\in\wedge^2 T_pM:\sigma\wedge\sigma=0,\|\sigma\|=1\}$. Hence $R$ could have some zero (or even negative) eigenvalues at the same time as the restriction $\sec(\sigma)=\langle R(\sigma),\sigma\rangle$ of its quadratic form to $Gr_2(T_pM)$ is positive. For example, in $\mathbb C P^n$ this is what happens: $R$ is positive-semidefinite (and has nontrivial kernel), but this kernel does not intersect the Grassmannian.

Algebraically, there is an intermediate curvature condition between $\sec>0$ and $R>0$ called strongly positive curvature, that might be of interest to you. Namely, a curvature operator $R\colon\wedge^2 V\to\wedge^2 V$ has strongly positive curvature if there exists a $4$-form $\omega\in\wedge^4V$ such that $R+\omega$ is positive-definite. Here, $\omega\in\wedge^4 V$ is identified with a symmetric endomorphism $\omega\colon\wedge^2V\to\wedge^2V$ via $$\langle\omega(\alpha),\beta\rangle=\langle\omega,\alpha\wedge\beta\rangle.$$ Clearly, $R>0$ implies strongly positive curvature (take $\omega=0$). Since $\sec(\sigma)=\langle R(\sigma),\sigma\rangle$ and $\langle\omega(\sigma),\sigma\rangle=\langle\omega,\sigma\wedge\sigma\rangle=0$ if $\sigma$ is decomposable, strongly positive curvature implies $\sec>0$. Together with R. Mendes, over the last years, I have pursued a systematic study of this curvature condition; see the following for details: https://arxiv.org/abs/1403.2117, https://arxiv.org/abs/1412.0039, https://arxiv.org/abs/1511.07899. For example, $(\mathbb C P^n,g_{FS})$ has strongly positive curvature, since $R+\varepsilon (\omega_{FS}\wedge\omega_{FS})>0$ for small $\varepsilon>0$, where $\omega_{FS}$ is the Fubini-Study $2$-form.

The upshot is that almost all known examples of manifolds with $\sec>0$ actually satisfy this much stronger condition. For manifolds with $\sec\geq0$, actually all known examples satisfy the analogous "strongly nonnegative curvature" condition (which requires $R+\omega$ to be positive-semidefinite).