fix: allow user with actions write permission to run a workflow from the web UI (#7245)

Users with permissions actions write can reschedule a workflow or start a new one over the API. But if the user try to use workflow dispatch on the web GUI it fails with 404 error.

Debug log says
```
router: completed POST /<owner>/<repo>/actions/manual for <client ip>, 404 Not Found in 21.6ms @ context/permission.go:17(web.registerRoutes.RequireRepoAdmin)
```

I think this is not correct, users with permissons action write should also be able to run a new workflow over the GUI. This can be fixed with this little change.

## Testing

- Create a repository in an organization with [a manual workflow](https://forgejo.org/docs/next/user/actions/#onworkflow_dispatch)
- Create a user and assign it action write permissions for this repository in a team
- Login as the user and verify they are allowed to run the workflow via the web UI

<!--start release-notes-assistant-->

## Release notes
<!--URL:https://codeberg.org/forgejo/forgejo-->
- User Interface bug fixes
  - [PR](https://codeberg.org/forgejo/forgejo/pulls/7245): <!--number 7245 --><!--line 0 --><!--description YWxsb3cgdXNlciB3aXRoIGFjdGlvbnMgd3JpdGUgcGVybWlzc2lvbiB0byBydW4gYSB3b3JrZmxvdyBmcm9tIHRoZSB3ZWIgVUk=-->allow user with actions write permission to run a workflow from the web UI<!--description-->
<!--end release-notes-assistant-->

Reviewed-on: https://codeberg.org/forgejo/forgejo/pulls/7245
Reviewed-by: Earl Warren <earl-warren@noreply.codeberg.org>
Co-authored-by: Stefan <stku@noreply.codeberg.org>
Co-committed-by: Stefan <stku@noreply.codeberg.org>

routers/web/web.go

@@ -1393,7 +1393,7 @@ func registerRoutes(m *web.Route) {
 			m.Get("", actions.List)
 			m.Post("/disable", reqRepoAdmin, actions.DisableWorkflowFile)
 			m.Post("/enable", reqRepoAdmin, actions.EnableWorkflowFile)
-			m.Post("/manual", reqRepoAdmin, actions.ManualRunWorkflow)
+			m.Post("/manual", reqRepoActionsWriter, actions.ManualRunWorkflow)
 
 			m.Group("/runs", func() {
 				m.Get("/latest", actions.ViewLatest)