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You probably meant $\beta_{ij}^k=\beta_{ji}^k$ (the cup-product is commutative). With that condition, there is always a simply-connected compact manifold $7$-manifold $M$ whose only non-trivial cohomology groups are $H^0(M)\cong\mathbb{Z}$, $H^2(M)\cong\mathbb{Z}^r$, and $H^0(M)\cong\mathbb{Z}^s$, and such that the cup-product is as you indicate.

The manifold $M$ can be constructed from a CW-complex $X$ defined as the mapping cone of a certain map $\varphi\colon\vee_s S^3\rightarrow\vee_rS^2$by the usual method of taking . There is a finite simplicial complex subpolyhedron $K$ with K\subset\mathbb{R}^7$ of the same (even simple) homotopy type of $X$, embedding $K$ in some euclidean space $\mathbb{R}^m$ in a piecewise linear way, and taking see http://math.berkeley.edu/~stall/embkloz.pdf. Take $M$ to be the closure of a certain nice regular neighbourhood of the embedding, which will be $K\subset\mathbb{R}^7$. This $M$ is a compact connected manifold, simply-connected $7$-manifold (with boundary in generalboundary). The manifold $M$ is (simply) homotopy equivalent to $X$.You can actually bound $m$ uniformly, as you wish, since only finite 4-dimensional CW-complexes are used, but I don't remember the bound.

Let me now indicate how to define $\varphi$ to get the desired cohomology ring, see Baues' book on 4-dimensional complexes.

Recall that $\pi_3(\vee_rS^2)\cong\Gamma(\mathbb{Z}^r)$. Here $\Gamma$ is Whitehead's functor, defined for any abelian group $A$ by the existence of a universal quadratic map $\gamma\colon A\rightarrow\Gamma(A)$ (not a homomorphism), i.e. a map such that $\gamma(a)=\gamma(-a)$ and the cross effect map $A\times A\rightarrow \Gamma(A)\colon (a,b)\mapsto (a|b)=\gamma(a+b)-\gamma(a)-\gamma(b)$ is bilinear. The isomorphism $\pi_3(\vee_rS^2)\cong\Gamma(\mathbb{Z}^r)$ is defined by the map $\mathbb{Z}^r\cong \pi_2(\vee_rS^2)\rightarrow \pi_3(\vee_rS^2)\colon f\mapsto f\eta$ given by pre-composition with the Hopf map $\eta\colon S^3\rightarrow S^2$, which is universal with those properties. The group $\Gamma(\mathbb{Z}^r)$ is free abelian of rank $\binom{r+1}{2}$. If $\{e_1,\dots,e_r\}\subset \mathbb{Z}^r$ is a basis, then $\{\gamma(e_1),\dots,\gamma(e_r)\}\cup\{(e_i|e_j) \,;\, 1\leq i < j \leq r\}\subset \Gamma(\mathbb{Z}^r)$ is a basis.

There is a natural homomorphism $\tau\colon \Gamma(A)\rightarrow A\otimes A$ defined by the quadratic map $A\rightarrow A\otimes A\colon a\mapsto a\otimes a$. If $A=\mathbb{Z}^r$ then $\tau(\gamma(e_i))=e_i\otimes e_i$ and $\tau(e_i|e_j)=e_i\otimes e_j+e_j\otimes e_i$.

A homotopy class $\varphi\colon\vee_s S^3\rightarrow\vee_rS^2$ is essentially the same as a homomorphism $f=\pi_3(\varphi)\colon \mathbb{Z}^s\rightarrow \Gamma(\mathbb{Z}^r)$. The cohomology of the mapping cone $X$ is obviously $H^0(X)\cong\mathbb{Z}$, $H^2(X)\cong\mathbb{Z}^r$, $H^4(X)\cong\mathbb{Z}^s$, and zero otherwise. The cup-product $H^2(X)\otimes H^2(X)\rightarrow H^4(X)$ turns out to be the $\mathbb{Z}$-linear dual of $\tau f\colon \mathbb{Z}^s\rightarrow \Gamma(\mathbb{Z}^r)\rightarrow \mathbb{Z}^r\otimes \mathbb{Z}^r$. Hence, if $\{e_1,\dots,e_r\}\subset \mathbb{Z}^r$ and $\{\bar e_1,\dots, \bar e_s\}\subset\mathbb{Z}^s$ are the canonical bases, given by the inclusions of the factors of the wedges, it is enough to define $f$, and hence $\varphi$, as $f(\bar e_k)=\sum_{i=1}^r \beta_{ii}^k\gamma(e_i)+\sum_{1\leq i<k \leq r}\beta_{ij}^k(e_i|e_j)$.
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You probably meant $\beta_{ij}^k=\beta_{ji}^k$ (the cup-product is commutative). With that condition, there is always a simply-connected compact manifold $M$ whose only non-trivial cohomology groups are $H^0(M)\cong\mathbb{Z}$, $H^2(M)\cong\mathbb{Z}^r$, and $H^0(M)\cong\mathbb{Z}^s$, and such that the cup-product is as you indicate.

The manifold $M$ can be constructed from a CW-complex $X$ defined as the mapping cone of a certain map $\varphi\colon\vee_s S^3\rightarrow\vee_rS^2$ by the usual method of taking a finite simplicial complex $K$ with the homotopy type of $X$, embedding $K$ in some euclidean space $\mathbb{R}^m$ in a piecewise linear way, and taking $M$ to be the closure of a certain regular neighbourhood of the embedding, which will be a compact connected manifold, with boundary in general. The manifold $M$ is homotopy equivalent to $X$. You can actually bound $m$ uniformly, as you wish, since only finite 4-dimensional CW-complexes are used, but I don't remember the bound.

Let me now indicate how to define $\varphi$ to get the desired cohomology ring, see Baues' book on 4-dimensional complexes.

Recall that $\pi_3(\vee_rS^2)\cong\Gamma(\mathbb{Z}^r)$. Here $\Gamma$ is Whitehead's functor, defined for any abelian group $A$ by the existence of a universal quadratic map $\gamma\colon A\rightarrow\Gamma(A)$ (not a homomorphism), i.e. a map such that $\gamma(a)=\gamma(-a)$ and the cross effect map $A\times A\rightarrow \Gamma(A)\colon (a,b)\mapsto (a|b)=\gamma(a+b)-\gamma(a)-\gamma(b)$ is bilinear. The isomorphism $\pi_3(\vee_rS^2)\cong\Gamma(\mathbb{Z}^r)$ is defined by the map $\mathbb{Z}^r\cong \pi_2(\vee_rS^2)\rightarrow \pi_3(\vee_rS^2)\colon f\mapsto f\eta$ given by pre-composition with the Hopf map $\eta\colon S^3\rightarrow S^2$, which is universal with those properties. The group $\Gamma(\mathbb{Z}^r)$ is free abelian of rank $\binom{r+1}{2}$. If $\{e_1,\dots,e_r\}\subset \mathbb{Z}^r$ is a basis, then $\{\gamma(e_1),\dots,\gamma(e_r)\}\cup\{(e_i|e_j) \,;\, 1\leq i < j \leq r\}\subset \Gamma(\mathbb{Z}^r)$ is a basis.

There is a natural homomorphism $\tau\colon \Gamma(A)\rightarrow A\otimes A$ defined by the quadratic map $A\rightarrow A\otimes A\colon a\mapsto a\otimes a$. If $A=\mathbb{Z}^r$ then $\tau(\gamma(e_i))=e_i\otimes e_i$ and $\tau(e_i|e_j)=e_i\otimes e_j+e_j\otimes e_i$.

A homotopy class $\varphi\colon\vee_s S^3\rightarrow\vee_rS^2$ is essentially the same as a homomorphism $f=\pi_3(\varphi)\colon \mathbb{Z}^s\rightarrow \Gamma(\mathbb{Z}^r)$. The cohomology of the mapping cone $X$ is obviously $H^0(X)\cong\mathbb{Z}$, $H^2(X)\cong\mathbb{Z}^r$, $H^4(X)\cong\mathbb{Z}^s$, and zero otherwise. The cup-product $H^2(X)\otimes H^2(X)\rightarrow H^4(X)$ turns out to be the $\mathbb{Z}$-linear dual of $\tau f\colon \mathbb{Z}^s\rightarrow \Gamma(\mathbb{Z}^r)\rightarrow \mathbb{Z}^r\otimes \mathbb{Z}^r$. Hence, if $\{e_1,\dots,e_r\}\subset \mathbb{Z}^r$ and $\{\bar e_1,\dots, \bar e_s\}\subset\mathbb{Z}^s$ are the canonical bases, given by the inclusions of the factors of the wedges, it is enough to define $f$, and hence $\varphi$, as $f(\bar e_k)=\sum_{i=1}^r \beta_{ii}^k\gamma(e_i)+\sum_{1\leq i<k \leq r}\beta_{ij}^k(e_i|e_j)$.